TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA
CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
CHAPTER-1 : BASICS OF NETWORKS
,
The most basic quantity in an electronic circuit is electric charge Q.
,
The charge of an electron is 1.602 × 10–19 C. One coulomb charge is
PERSONAL REMARK :




defined as the charge possessed by 
 electrons.
–19
 .  10 
Therefore, 1 coulomb charge = charge of 6.24 × 1018 electrons.
,
The total charge of an isolated system remains constant regardless
of changes within the system itself.
,
1 A current = flow of 6.24 × 1018 electrons per second through an
area.
,
The potential of a point is 1 volt is 1 joule of work done in bringing a
1-coulomb charge from infinity to that point. However, the voltage
Vab between two points a and b is work, W required to move a unit
positive charge from a to b.
Vab =
dW
dq

Difference Between Network and Circuit
,
Any combination and interconnection of network elements like R, L,
C or electrical energy sources are known as a NETWORK.
,
However, a closed energized network is known as a CIRCUIT.
,
It may be noted that network need not contain an energy source, but
a circuit must contain at least one energy source
Note : All the circuits are networks, but vice-versa is not true.

Difference Between Mesh and Loop
,
A loop denotes a closed path obtained by starting at node and returning
back to the same node.
,
A mesh is a smallest closed path which does not contain any closed
path inside it or we can say mesh does not contain any other loop
within it.
Note: All the mesh are loop, but vice versa is not true.

Difference Between Lumped & Distributed Network
,
Lumped Network
Lumped network are those in which we can separate resistors,
capacitors and inductors physically.
,
Distributed Network
A circuit in which the voltage and current are functions of time and
position is called a distributed parameter circuit. While a circuit in
which the voltage and current are functions of time only is called
lumped parameter circuit. In distributed network, resistors,
capacitors and inductors cannot be electrically separated and isolated
as separate elements. For example, transmission line.
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
BASIC CIRCUIT ELEMENT
Active Element
Note: (i)
Current
Source
Non-linear Passive
Element Example
-Diode and transistors
Resistors, Inductors and Capacitors may be linear or non-linear,
while diodes and transistors are alway non-linear
(iii)
A circuit / Network element is linear if the relation between the
current and voltage involves a constant coefficient.
(iv)
v–i relationship of resistor(with linear coefficient of resistor i.e.  =0),
& inductor and capacitor [ with zero initial conditions i.e.
iL(0-)=0 A and vC(0-)=0 V] are linear
1
di
i dt )
,v=
C
dt

Linear Element : A circuit element is said to be linear if the relation
between current and voltage involves a constant coefficient, e.g.
V = IR ; V = L

Linear Passive
Element. Example
-R, L, C
(ii)
(v)

It does not generate Electricity
but either consumes it or stores it.
Most active elements are independent of other circuit variables, but
some elements are dependent (modeling elements such as transistors
and op-amps would require dependent sources)
(v = i R, V = L
,

Passive Element
Capable of Generating
Electrical energy
Voltage Source
(Such as Battery
or generator)
PERSONAL REMARK :
di
1
i dt
, V=
dt
C

A linear network is one in which principle of superposition (i.e.
additivity and homogeneity or scaling) holds. A non-linear
network is one which contains non-linear elements.
Additivity : If the input signals x1(t) and x2(t) corresponds to the
output signals y1(t) and y2(t), respectively then the input signal
{x 1 (t) + x 2 (t)} should corresponds to the output signal
{y1(t) + y2(t)}.
Homogeneity or Scaling: If the input signal x1(t) corresponds to
the output signal y1(t) then the input signal a1x1(t) should corresponds
to the output signal a1y1(t) for any constants a1.
Note : For linear there must be constant slope from – to + in the v – i
characterstics plot.
Note: 1.
When element properties and characteristics are independent of direction of
current than the element is called as bidirectional or bilateral element. For
example R, L and C are bilateral element.
R
R
I'
I
V
V
If | I | = | I' | then element is called bilateral elelment.
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
2.
A linear network is passive if -
(a)
The energy delivered to the network is non-negative for any orbitarary
excitiation and
(b)
if no voltages or currents appear between any two terminals before an
excitation is applied.
(c)
When V/I ratio is positive in both co-ordinates then the element is passive
otherwise active.
PERSONAL REMARK : 
Ex. The v-i characteristic of an
element is shown in the figure
given below. The element is
Note : If the magnitude of power is positive it means power is delivered to the system (i.e.
system or element is passive) whereas if the magnitude of power is negative
it means power is given out by the system (i.e. system or element is active).
3.
v
When element properties and characteristics depends on direction of the
current then the element is called as unidirectional or unilateral element
for example diode is a unilateral element.
i
(a) non-linear, active, non-bilateral
I
I'
(b) linear, active, non-bilateral
(c) non-linear, passive, non-bilateral
V
V
Here | I |


(d) non-linear, active, bilateral
| I' |
4.
Every linear element must obey the bilateral property. However, vice-versa
is not necessarily true.
5.
It may be noted that bilateral curve identical in opposite-plane not in adjacent
plane.
Sol. The given figure is 
because the relation
between the current and
voltage does not involves a
constant coefficient.
Various Possible V-I Characteristics of an Element
I
(a)


Non-linear

Unilateral (or Non-Bilateral)

Passive

Non-linear

Bidirectional

Passive

linear

Bilateral

Passive

Non-linear

Unilateral

Active

Linear

Bilateral

Active

Non-linear

Unilateral

Active

(b)

V
I
(c)

–V
V
I
(d)

V
V
(e)

I
V
(f)

I
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Active
Because the ratio between
the current and voltage is
both positive and negative.
V
I
Non-linear

Unilateral/Non-Bilateral
Because v is positive when
i is changes from positive
to negative.
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)

RESISTOR
Resistor is a two terminal Electrical / Electronic component that
resists an electric current by producing a voltage drop between its
terminals in accordence with ohm's law, R =
,
V
I
Instantaneous power absorbed in the resistor is given by
p = vi = i R × i = i2 R (watts)
,
Energy converted into heat energy is given by
W=
,

t
0
p dt 

t
0
i2R dt = i2 Rt (Joules)
Table given below shows the colour coding of Four-Band Arial
Resistors
Colour Ist band 2nd band 3rd band 4th band
Temp
(multiplier) (tolerance)
Black
Brown
0
1
0
1
× 10º
× 101
–
 1% (F)
–
100 ppm
Red
2
2
× 1021
 2% (G)
50 ppm
Orange
3
3
× 10
3
–15 ppm
4
–25 ppm
Yellow
4
4
× 10
Green
5
5
× 105
Blue
6
6
× 106
Violet
7
7
× 107
Grey
8
8
× 108
9
White
Gold
9
–
9
–
× 10
× 0.1
Silver
–
–
× 101
None
–
–
× 0.01
 0.5% (D)
 0.25% (C)
–
 0.1% (B)
 0.05%(A)
–
 5% (J)
–
 10% (K)
 20% (M)
–
PERSONAL REMARK : 
Ex.1 A voltage wave having a
time variation of 20 V /sec is
applied to a pure capacitor
having a value of 25  F..
Find (a) the current during the
period 0 < t < 1 sec. (b) charge
accumulated across the
capacitor at t = 1 sec, (c) power
in the capacitor at t = 1 sec, and
energy stored in the capacitor at
t = 1 sec.
Sol. (a)Current through the
capacitor, i may be expressed as
iC
dv
 25  106  20
dt
 500  A
(b) At t = 1 sec, v = 20 V.
Charge q at t =1sec may be
expressed as
q  Cv  25  106  20
 500  C
(c) At t =1sec, power
–
p  v  i  20  500  106
–
 1  10 2 W
–
–
(d) At t =1sec, energy stored in the
capacitor, WC, can be expressed
1
1
WC  Cv2   25 106  (20)2
2
2
–
 5  10 3 J
Note: Red to violet are the colors of the rainbow where red is low energy and violet
is high energy.
,
As an example, let us take a resistor which (read left to right) displays
the colors yellow, violet, yellow, brown. We take the rist two bands
as the value, giving us 4, 7. Then the third band, another yellow,
gives us the multiplier 104. Our total value is then 47 × 104 , totaling
470.000or 470 k. Our brown is then a tolerance of  1%.
,
Resistors use specific values, which are determined by their tolerance.
These values repeat for very exponent; 6.8, 68, 680, and so forth.
This is useful because the digits, and hence the first two or three
stripes, will always be similar patterns of colors, which make them
easier to recognize.
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Ex.2 The time rate of change of a
voltage applied across a 1 F
capacitor is 2 V/s. This means
that the current flowing
through the capacitor is __
(a) 2 X 10-6A
(b) 2 A
-6
(c) 0.5 X 10 A (d) 0.5 A
(JTO-EC-2009)
Sol.(a)
Given,C = 1 F and
dVC
= 2 V/s,
dt
ic= ? .We Know that current
through capacitor is given by
dV
relation, ic = C C
dt
-6
= 1 x 10 x 2 = 2 x 10-6 A
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)

CAPACITOR
Capacitor is an electrical device that can store energy in the electric
field between pair of closely spaced conductors. When a current is
applied to the capacitor, electronic charges of equal magnitude, but
of opposite polarity, build up on each plate.
,
Capacitance is the property of material by virtue of which it opposes
the variation in potential between the two sides
C=
,
PERSONAL REMARK : 
Ex. The equivalent capacitance
across ‘ab’ will be
Q
V
(b)
0.1F
(c) 0.5F
(d)
0
a
where, Q = charge, V = potential
The capacitrance may also be defined as, C =
(a) 0.2F
b
0rA
d
(ISRO-EC-2010)
where, C = Capacitance is proportional to the dielectric and area of
the plates, and is inversely proportional to the distance between the
plates.
Sol.(b)The circuit can be redrawn
c
as (x )
(x )
1
3
a
0 = 8.85 × 10– 12 C2N/m2, r = Relative permittivity
A = Area of the plate, d = Spacing between two plates
,
,
(x 4)
(x2 )
d
Since all the capacitors are same
When n capacitances are connected in series, the equivalent
capacitance is given by
1
1
1
1
1
=
+
+
+ .... +
C eq
C1
C2
C3
Cn
b
x1 x 3
is same

x2 x4
So, no current flows through c-d
The circuit look as
When n capacitances are connected in parallel, the equivalent
capacitance is given by
a
b
Ceq = C1 + C2 + C3 + ..... + Cn
The circuit is simplified to
SOME CONSERVATION LAWS
(a)
The conservation of charge
b
a
q1 = q2 and C1 V1 = C2 V2 ; i  
(b)
The conservation of flux linkage
1 = 2 and L1 i1 = L2 i2 ; v  
(c)
So, C a  b  0.05  0.05F  0.1F
The conservation of momentum
Ex.
P1 = P2 and m1 v1 = m2 v2 ; F  
,
From relation Q = CV, the current, i =
dQ
dV
dC
 C
 V
dt
dt
dt
In most physical cases the capacitance is constant with time.
dV
therefore, i = C
....(A)
dt
From equation (A) it is clear that for an abrupt change of voltage
across the capacitor, the current becomes infinite.
,
Voltage across the capacitor is given by
VC (t) =
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t

–
i c (t) dt 
1
C
0

–
i c (t) dt 
The total capacitance of two
capacitors is 25 F when
connected in parallel and 4 F
when connected in series.
The individual capacitances
of the two capacitors are
(a) 1 F and 24 F
(b) 3 F and 21 F
(c) 5 F and 20 F
(d) 10 F and 15 F
Sol.(c)
1
C
t
 i c (t) dt

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V C (t) = VC (0–) +
or
1
i c (t) dt
C

Where, VC (0–) is the initial voltage across the capacitor.
,
For the zero initial voltage, VC(t) =
1
C
t
 i c (t) dt
..... (B)

From equation (B) it is observed that for a finite change of current in
zero time, the integral must be zero. Therefore, voltage across a
capacitor cannot change instantaneously i.e. VC (0–) = VC (0+) if i(t)
is not a (t) or derivative of (t)

INDUCTOR
An electric current i flowing round a circuit produces a magnetic
field and hence a magnetic flux  through the circuit. The ratio of
magnetic flux to the current is called the inductance or self inductance,

i.e. L =
i
,
The relation with induced voltage, v and inductance L
di L
d
=N
dt
dt
v=L
..... (C)
From equation (C) it is clear that for an abrupt change in
current, the voltage across the inductor becomes infinite.
,
Inductance
:
is
the
property
of
conductor
(or coil) by virtue of which it opposes any change in direction or
magnitude of current flowing through itself. It is given by
L=
N
I
henry
Where, N = No. of turns in the coil,  = Flux set by current I.
 0 ur N2 A
l
,
Also,L =
,
Current through inductor is given by
i (t) =

L
t
–V dt
or iL (t) = iL (0–) +
henry

L
or

L

L

– VL(t) dt +
t

VL (t) dt
PERSONAL REMARK : 
Ex. Consider a current source
i(t) connected across a 0.5 mH
inductor, where i(t) = 0 A for
t < 0 and
i(t) = (8e–250t – 4e–1000t) A
for t  0 . The voltage across the
inductor at t = 0 s is
(ISRO-EC-2006)
(a) 0.5 V
(b) 1 V
(c) 2 V
(d) 4 V
Sol.(b) Voltage across the inductor
is given by , VL  L
di L
dt
di L
 8( 250)e  250t  4(1000)e 1000 t
dt
di L
 2000e  250t  4000e 1000t
dt
At , t = 0
di L
 2000  4000  2000
dt
di L (0  )
 0.5  10 3  2000
dt
Therefore, VL = 1 V
VL  L
Ex. A current having a variation
shown in figure is applied to a
pure inductor having a value of
2 H. Calculate the voltage
across the inductor at time
t = 1 and t = 3 sec.
i, amperes
10
t
 iL (t) dt
,
When n inductors are connected in series the equivalent inductance
is , Leq = L1 + L2 + L3 + .....+ Ln
,
When n inductors are connected in parallel the equivalent inductance,
1
1
1
1
1
=
+
+
+ .... +
L eq
L1
L2
L3
Ln
0
1
2
3
t, sec
Sol. For the period 0 < t < 1 sec
Current i = 10 t A
Rate of change of current
.
di
 10A / sec
dt
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
,
When initial current is zero
iL(t) =
 t
VL (t) dt
L 
....(D)
From equation (D) it is observed that for a finite change in voltage in
zero time the integral must be zero. Therefore, the current through
inductor cannot change instantanously i.e. i L(0–) = iL(0+).
,
Explain why a capacitor is considered as a linear clrcuit
element.
Let VC and VC2 individually excite a relaxed capacitor producing
1
the respective currents.
dVC1
i C1 = C
dt
and i C 2 = C
dVC 2
dt
iC = C
dVC1
dVC 2
d
( VC + V ) = C
+C
= ( i C1 + i C 2 )
C2
1
dt
dt
dt
This shows that the v–i characteristic of a capacitor obeys the
superposition principles. Therefore, a capacitor is considered as a
linear element.
,
Explain why an Inductor is considered as a linear circuit
element.
Let VL1 and VL 2 individually excite a relaxed Inductor producing
the respective currents,
1
iL1 =
L

1
VL1 dt and iL2 =
L
1
L
 VL dt 
1
L
 VL
2
dt
 VL
1


 VL 2 dt  i L1  i L 2

If an impulse voltage is applied to an inductor what will be the
result
1
iL =
L
t

–
 (t – T) dt 
For the period 1 < t < 3 sec
Rate of change of current
di
 5 A /sec
dt
Therefore, at t = 3 sec, voltage
across the inductor is
di
 2  5  10 V
dt
Ex. A voltage wave having the
time variation shown in figure
is applied to a pure inductor
having a value of
0.5 H.
Calculate the current through the
inductor at times t = 1, 2, 3, 4, 5
sec. Sketch the variation of
current through the inductor over
5 sec.
v
volts
10
1 2
3 4
5
t
sec
10
This shows that the v – i characteristic of an inductor obeys the
superposition principles.Therefore an inductor is considered as a linear
element.
,
di
 2  10  20 V
dt
0
Let iL be the current induced by a voltage ( VL + V )
L2
1
 iL1 =
L
L
Let iC be the current induced by a voltage ( VC + VC )
1
2

PERSONAL REMARK : 
Therefore, at t = 1 sec, voltage
across the inductor is
1
u (t – T)
L
Sol. For the period 0 < t < 1 sec,
v = 10 V; i(0) = 0.
The current i may be expressed
as , i 

I t
vdt  i(0)
L 0
t
1 t
10d

20
0 dt  20t
0.5 0
Then at t = 1 sec, we get
i = 20  1 = 20 A
For the period 1 < t < 3 sec,
v =  10 V ; i(1) = 20 A,
then current
Thus, an impluse voltage applied to an inductor L results
instantaneously in a current of 1/L.
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However, we know that the current in an inductor cannot change
instantaneously. Here, the instantaneous current generated is an
unusual behaviour of an inductor and this happens because of the
fact that the driving voltage in the form of an impulse is also an
unusual voltage.
,
If a unit impulse current is applied to a capacitor what will be
the result ?
t
1
The voltage-current relationship of capacitor is VC =
i(t) dt
C –
If an impulse current is applied to a capacitor then the resulting
voltage across the capacitor is given by
t
1
VC =
C

 (t – T) dt 
–
1
u (t – T)
C
i
TYPES OF ENERGY SOURCES
Energy Sources

1
vdt  i(1)
L 1
t
t
1 t
10dt  20  20 dt  20

1
0.5 1
=  20(t  1) + 20
Then at t = 2 sec, we get
i =  20  (2  1) + 20
=  20 + 20 = 0 A
And at t = 3 sec, we get
i = 20  (3  1) + 20 =  40 + 20
=  20 A
For the period 3 < t < 5 sec,
v = 10V, i(3) =  20 A

i
Thus, an impulse current applied to a capacitor C results in an
instantaneous voltage of 1/C.
,
PERSONAL REMARK :
1 t
vdt  i(3)
L 3
t
1 t
10dt

20

20
3 dt  20
0.5 3
= 20 (t  3)  20

i, amperes
20
Independent
Energy Sources
1
Dependent/Controlled
Energy Sources
Voltage Controlled
Voltage Source
(VCVS)
Voltage Source
V S +–
Voltage Controlled
Current Source
(VCCS)
KV
Current Controlled
Current Source
(CC CS)
KI
RS=0
RS=0
+
VL
–
Load VS
±
+
VL
–
Load
Current Controlled
Voltage Source
Practical
Ideal
Current Source Current Source (CC VS)
VL = VS– I RS
VL = VS
IL
IS
V
RS=
+
– KI
IL
Load IS
RS=
I
4
0
5
t, sec
IL= IS–
IL= IS
Figure : Variation of current through
the inductor
Then at t = 4 sec, we get
i = 20  (4  3)  20 = 20  20 = 0 A
And at t = 5 sec, we get
i = 20  (5  3)  20 = 40  20 = 20A
Ex.2 A current having variation
shown in figure is applied to a
pure capacitor having a value
of 500 F. Calculate the
charge, voltage, power and
energy at time t = 1 sec.
Load
V
I
3
20
Current Source
Practical
Voltage Source
Ideal
Voltage Source
+
– KV
2
VL
RS
V
V
i, amperes
100 mA
I
I
Note : (i) Dotted lines shows ideal response (ii)
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Bold line shows paractical Response
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
IDEAL VOLTAGE AND CURRENT SOURCE
,
An independent ideal voltage source has the following features

It is a voltage generator whose output voltage remains
absolutely constant whatever be the value of the output current.
or in otherwords we can say output voltage i.e. VL must be
constant equal to VS irrespective of the load.
PERSONAL REMARK : 
Sol. For period 0 < t < 1 sec,
i  100  103 t = 0.1t A
At t = 1 sec.
t
t
q   idt   0.1t dt
0


Note: 

,
It has has zero internal resistance (i.e. RS = 0) so that voltage
drop in the source is zero.
The power drawn by the source is zero.
An ideal voltage source is not practically possible. No voltage source can
maintain its terminal voltage constant even when its terminal are short
circuited.
Lead-acid battery and a dry-cell are some examples of constant voltage
source which can produce constant terminal voltage within a specified range
of output current.
0
t 1
 t2 
 0.1     0.05[t 2 ]tt 10
 2  t0
 0.05[1  0]  0.05C
q 1 t
0.05t 2
v    0.1tdt 
C C 0
500  106
An independent ideal current source has the following features
= 100t2 = 100 V, where t = 1 sec
p = v  i = 100  0.1 = 10 W
It produces a constant current irrespective of the value of the
voltage across it.
WC   vidt   100t 2  0.01tdt



It has infinite internal resistance (i.e. RS =  )
It is capable of supplying infinite power.
Note: 
An Ideal independent current source is not practically possible. No current
source can maintain constant current even when its terminals are open
circuited. The output current (IL) of a practical current source decreases as
the output voltage increases.

A solar cell, which can produce constant current within a specified range of
output voltage.

t
t
0
0
t 1
 t4 
1
  t dt     [1  0]
0
 4  t 0 4
t
3
= 0.25 J
For the period 1 < t < 2 sec,
i = 0.2  0.1 t A At t = 2 sec
i = 0.2 A

Charge, q = qt = 1 +
A natural lightning can be considered to be an ideal current source.
t
0
id t
2
 0.05   (0.2  0.1t)dt
1
DEPENDENT ENERGY SOURCE
,
(i)
+
I1 = 0
+
– V2
(ii)
= 0.05 + 0.05 = 0.1 C
V2
–
–
Voltage Dependent Current Source (VCCS)
+
I1 = 0
V1
–
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I2
gmV 1

t2 
 0.05   0.2t  0.1  
2  t 1

= 0.05 + [0.2(2  1)  0.05(22  12)]
+
I2
V1
,
t2
Voltage Controlled/Dependent Voltage Source (VCVS)
+
V2
–
Classroom & Online Test Series
Foundation Batches also for 2nd & 3rd Year sturdents
Ex. In electronic circuits, for blocking
the DC component of a voltage
signal, a/an ______ is
connected in series with the
voltage source.
[SSC-JE : 2013]
(a) capacitor (b) diode
(c) resistor
(d) inductor
Sol.(a) Capacitor
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TECHGURU CLASSES for SSC-JE/RRB/UP-PCL/UPRVNL/ISRO/DRDO/TTA
CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
,
(iii)
Current Dependent Current Source (CCCS)
I2
+
+
I1
V2
 I1
V1= 0
(iv)
Current Dependent Voltage Source (CCVS)
I1
I2= 0
+
– RmI1
V1= 0
,
(a) –1/3 A
(b) 1/5 A
(c) 2/5 A
(d) 1 A
(DRDO-EE-2010)
Sol.(c)Using source transfromation
the circuit is redrawn as
V2
–
–
,
2A
i
+
+
,
3 – i1
i1
1A
–
–
,
PERSONAL REMARK : 
Ex. The current i thorugh the 1
resistor shown in figure is
+
–
2V +
–
Limitatations of Ohm's Law

It is not applicable to non-linear circuits like circuits with powered carbon,
thyrite etc.

It is not applicable to unilateral circuits, like circuits with electron tubes,
transistors etc.
Kirchhof's Voltage Law (KVL) is a consequence of the law of
conservation of energy, voltage being the energy (or work) per
unit charge.
Kirchhof's Current Law (KCL) is the consequence of law of
conservation of charge. Since the algebraic summation of the
charge must be zero, the time derivative of this summation must also
equal to zero.
SOME SPECIAL EQUIVALENT CIRCUIT
It may be always remember that voltage source cannot be connected
in parallel unless the two sources have identical voltages, and similarly,
the current sources cannot be connected in series unless identical.
The paralleling of generators with non-similar voltage waveforms,
results in heavy currents and equipment can damage.
4V
i
Applying KVL in the circuit
5i + 2 – 4 = 0 or i 
2
A
5
Ex. In the circuit shown below, the
current through the 3/11 
resistance between terminals A
and B is
(JTO-EE-2009)
(a) 4 A
(b) 1 A
(c) 2 A
(d) 5 A
1V
2V
3V
A
(a)
+
–
V1
+
–
V2
B
Sol. Applying KCL,at A we get
+
–
V1 V2
VA –3 VA –2 VA –1 VA



0
3
2
1
3/11
 1 1 1 11  3 2 1
VA  + +    + +
3 2 1 3  3 2 1
 2 + 3 + 6  22 
VA 
  3
6

(b)
i1
i2
i1 i 2
or VA = 18/33 V and
I(3/11
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
PERSONAL REMARK :
+
–
(c)
+
–
V1
+
–
V2
i1
i1 = i 2
i2
,
(e) If any element (except voltage
source) is connected in parallel
with voltage source than for
circuit / network analysis the
equivalent is shown.
V1 = V2
+
–
V +
V
Three equivalent networks shown below illustrating a procedure for
a source to be ‘Pushed through a node’.
R1
Vi
R2
L1
+
–
R1
Vi
R2
L1
+
–
R1
V1
+
–
L1
+
+
–
(f) If any element (except current
source) is connected in series with
current source than for circuit /
network analysis the equivalent is
shown.
R2
V1
Vi
I
(1)
(2)
i1
I
(3)
R1
i1
R1
R2
i2
R2
Ex. The Voltage across last resistor
is V. All resistors are unity.
Then VS is given by
+
+
V
–
VS
Figure : Two equivalent network
b
c
b
V1 – + a
e
c
– + – + V1
a
d
e
–+
V1
d
–
(a) 13 V (b) 8 V
(c) 4 V (d) None of these
(ISRO-EC-2006)
Sol.(a)From the circuit given
V1
+
VS
V2
+
V
–
I3
I1
–
Voltage across 1 resistance is V..
V
V A
1
Now, V2 = V + V = 2V
Apply nodal at V2
So, I3 
I 2  V2 
Figure : Two equivalent network
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2
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)

PERSONAL REMARK :
STAR TO DELTA TRANSFORMATION
A
Now,V1  I2 1 V2  3V  2 V  5 V
A
RC
RB
Apply nodal at node Vi
RAB
RA
O
RCA
B
I1 = V1 + V1 – V2 = 2 V1 – V2
or I1 = 10 A – 2 A = 8 A
C
RBC
C
B
Therefore,
VS  I1  1  V1  8 V  5 V
RA RB + RB RC + RC RA
RC
RAB =
or
or I2 = 3 V
RAB = RA + RB +
or VS  13 V
Ex. For
the
Delta-Wye
transformation as in Fig. shown,
the value of the resistance R is
RA RB
RC
Similarly,
(ISRO-EC-2007)
RBRC + RCR A + R ARB
RA
RBC =
R
and
R
R CR A + R A RB + RBR C
RB
RCA =
R
DELTA TO STAR TRANSFORMATION
A
A
RAB
RCA
B
RB
(a)
1

3
(b)
2

3
(c)
3

2
(d)
3
RA
O
RC
C
RBC
C
B
Sol.(a) Applying Delta ( ) –Wye
(Y) trasformation
RAB  RAC
RA = R + R + R
AB
BC
CA
RA 
Similarly,
Ex.
RAC  RBC
and
,
RAB  RBC
RB = R + R + R
AB
BC
CA
RC = R + R + R
AB
BC
CA
Resistance of a wire depends on its material and its size. It is given
by
R=
Where,
 = Resistivity of a material in cm
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3R
3R
RR
3R
So, equivalent delta resistance
= 15 each.
A = Area of cross-section of wire
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Three resistance 5 each are
connected in star. Values of
equivalent delta resistances
are
[SSC-JE : 2012]
(a) 1.5 each (b) 2.5 each
(c) 5/3 each (d) 15 each
Sol.(d)
l
A
l = Length of the wire, and
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RAB  RAC
11 1

 
RAB  RBC  RCA 11 1 3
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
PERSONAL REMARK :

NETWORK ANALYSIS
NODE
Analysis
MESH/LOOP
Analysis
No. of independent
equations required
=b–N+1
where, b = No. of Branches
N = No. of nodes
Note:
,
No. of independent equation
required = N – 1
Where, N = no. of nodes

Mesh  KVL + ohm's law

Supermesh  KVL + KCL + ohm's law

Nodal  KCL + ohm's law

Supernode  KCL + KVL + ohm's law
Choice between Loop and Node methods of analyzing a
network

The mesh method is generally used for circuits having many
series-connected elements, voltage sources or supermeshes
on the otherhand, nodal analysis is more suitable for networks
for circuits having many parallel-connected elements, current
sources or supernodes.

Usually if node voltages are required, nodal analysis is used;
if branch/mesh currents are required, loop analysis is used.
Note : However, there are some particular circuits where only one method can be
applied. For example, in the analyzing transistors circuits, mesh method is the
only possible method; while for op-amp circuits and for non-planar networks,
node method is the only possible method.
SOLVED EXAMPLES
1.
Find i1, i2 and V in the circuit shown figure.
+V
i1
3
i2
5A
4
2A
8A
0A
Sol. Assuming node A, B & C in the given circuit
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Applying KCL at node(A), we get
i1 C
+V
i2 + 2 A = 0 + 8 or i2 = 6 A
B
Ex. Find the value of V 0 in the
figure.
(DRDO-EE-2010)
5A
Applying KVL in mesh we get,
V  (3  6) +(4  2) = 0
4
3
i2
2A
or V = 10 V
8A

PERSONAL REMARK :
1A
0A
A
Applying KCL at node C we get, i1 + ix = i2 or i1 = 6  3 = 3 A
+
V0
Ex.2 A copper wire has a resistance of 0.85 Wat 200C. What will be its
resistance at 400C? Temperature coefficient of resistance of copper
at 00C is 0.004/0C.
1V +–
–
+
–
1V
[SSC-JE:EE-2014]
Sol. Given , R0 = 0.85 W, = 0.004/0C and T = 400C – 200C = 200C
We know that , RT = R0 (1+ T) = 0.85 (1+ 0.004  20) = 0.918 W
(a) 0.5 V
(b) –1/2
(c)
(d) –3/2
Sol.(a) Applying KCL at node V1
Ex.3 In the circuit shown below.What is the value of V?
5V
+
+V
0V
1A
V1
6V +
+
[SSC-JE:EE-2014]
V0
1V +–
–
+
–
1V
Sol. Applying KVL in the given loop , we get 6 – V – 5 = 0 or V = 1 Volt.
Ex.4 In figure 3, find the value of resistance R.
[SSC-JE : EE-2014]
1 + 1 – V1 = V1 – 1
or 2 V1 = 3 or V1 = 1.5
10 
Therefore, V0 = 1.5 – 1 = 0.5 V
2A
100 V
+
10 
R
Sol. Applying KCL in the node,we get
2V A
VA – 1 0 0 VA
 8 or VA = 40 Volts.

 2  0 or
10
10
10
and R =
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= 20 W
I
2
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
,
SUPERMESH ANALYSIS


PERSONAL REMARK :
Suppose any of the branches in the network has a current
source, then it slightly difficult to apply mesh analysis straight
forward. One way to overcome this difficulty is by applying
the supermesh technique.
A supermesh is constituted by two adjacent loops that have a
common current source. For the circuit, shown below.
R1
V +
–
Ex.
Consider the following
statements :
Any element connected in
1. series with a voltage source
is redundant
2. parallel with a voltage
source is redundant
R2
I1
R3
I2
I
(1)
(2)
I3
3. series with a current source
R4
is redundant
(3)
4. parallel with a current
Supermesh
source is redundant
Considering mesh (1) and (2) :
The correct statements are
V = R1I1 + R3 (I2 – I3)
or

V = R1I1 + R3 I2 – R3 I3
..... (i)
Considering mesh (3) : R3 (I3 – I2) + R3 I3 = 0
..... (ii)
(a) 1 and 3 (b) 2 and 3
(c) 3 and 4 (d) 1 and 2
Sol.(b)
Finally the current I from current source is equal to the difference
between two mesh currents i.e.I1 – I2 = I
We have, thus, formed three mesh equation which we can solve for
the three unknown currents in the network.
SOLVED EXAMPLES
Ex.1 Determine the current in the 5  resistor in the network shown
below :
(2)
(1)
–
50V +
10
I2
2
A
I1
3
2A
5
I3
1
(3)
Sol. Applying KVL in loop 1, 2 and 3
50 = 10 (I1 – I2) + 5 (I1 – I3)
10 (I2 – I1) + 2 I2 + I3 + 5 (I3 – I1) = 0
I2 – I3 = 2
Solving equation (i) , (ii) & (iii) , we get
I5 = I1 – I3 = 4.67 A
..... (i)
..... (ii)
..... (iii)
ALTERNATIVE METHOD
KCL at node A : 2 =
Now,
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VA – 50
V

 A or VA =
10


VA
70

= 4.67 A
5
3 5
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
,
SUPERNODE ANALYSIS
PERSONAL REMARK :
Suppose any of the branches in the network has a voltage source,
then it is slightly difficult to apply nodal analysis. One way to overcome
this difficulty is to apply the supernode technique.
For the circuit arrangement shown below :
Ex.
When KCL is applied at the
supernode in the circuit
shown, the current equation in
terms of node voltage V1 and
V2 is


–6 =
(b)
4=
V1 – V2 V1 – V2

2
20
(c)
4=
V1 V1 – V2

2
20
..... (i)
Due to the presence of voltage source Vy at node B and C. It is
slightly difficult to find out the current. Thus by using the concept of
supernode, we get
VB – VC = Vy
..... (ii)
V1 V2
+
2
4
Sol.(a)Applying nodal using
supernode concept, we get
..... (iii)
V1 V2

 10 – 4  0
2
4
Writing KCL equation for nodes B and C, we get
V – VX
V
VB – VA
V
 B  C
 C =0
R2
R3
R4
R5
V1 V2

2
4
(a)
Writtng KCL equation at node A, we get
VA
VA – VB
Is = R 
R2
1

From the above three equations, we can find the three unknown
voltages.
(d)
or
4=
V1 V2

=–6
2
4
Ex.2 Determine the current in the 5  resistor for the circuit shown below
:
V1
V2
2
10A
3
+–
V3
20V
1
5
2
10V
Sol. At node 1, we get
V1
V – V2
 1
= 10
3
2
..... (i)
Using supernode concept, we get
V2 – V3 = 20
..... (ii)
KCL at node 2 and 3, we get
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
PERSONAL REMARK :
V – 10
V
V2 – V1
V
 2  3
 3 =0
2
1
5
2
..... (iii)

Solving equation (i), (ii) and (iii), we get,
V3 = – 8.42 V
V2 = 20 + V3 = 20 – 8.42 = 11.58 V

Current I5 =
V3 – 10
– 8.42 – 10
–
= – 3.68 A
5
5
(i.e. current passes towards node 3)
Ex.
Find the current Ix which flows through the 3 resistor in the
circuit of the figure below :
1
6V
V1
V2 2 
Q
–
+
12 
–
3
3
6V
+
168 V +–
Sol. Circuit can be modified using source transformation as :
1
6V
2
+ –
14A
3
12
3
Ix
–
+
6V
From the above circuit we get V1 – V2 = 6  V2 = V1 – 6.... (1)
Applying Supernode analysis
V1 – 168 V1 V2
V 6


 2
=0
13
13
3
2
or
V1 – 168 V1 V1 – 6 V1



=0
13
13
3
2
or 6V1 – 6 × 168 + 26 V1 + 26 V1
– 26 × 6 + 39 V1 = 0
or 97 V1 = 1008 + 156
or 97 V1 = 1169 or V1 = 12 V
and Ix =
V1
12

=4A
3
3
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
RULES FOR CALCULATING ABSORBED/DELIVERED
POWER
,
PERSONAL REMARK :

The power absorbed by any circuit element as shown below with
terminals labelled A and B is equal to the voltage drop from A to B
multiplied by the current through the element from A to B, i.e.
P = VAB. IAB.
IAB
+
,
–
B
A
The element X
I
+
+
I
V
X
V
X
–
–
Absorbing Power
Delivering Power
Pabsorbed = –VI
P delivered = VI
,
Pabsorbed = VI
Pdelivered = –VI
Note : Consider the following possible situations -
I
I
+
–
+
–
V
+
I
V
V
+
I
V
–
Delivering
Power
Absorbing
Power
–
Delivering
Power
Absorbing
Power
SOLVED EXAMPLES
Ex.1 Compute the power absorbed by each of the elements as shown
below.
–1A
(a)
+
–3A
–
40V
(b)
+
20V
–
4A
(c)
Sol. (a)
–
+
10V
P = –40 × (–1) = 40 W
(b)
P = 20 × (– 3) = – 60 W
(c)
P = 10 × 4 = 40 W
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Ex.2 Find the power of each element for the circuit
PERSONAL REMARK :

IA
5
20 V
3IA
5V
Sol. Let the voltage at node A is VA therefore VA = 20 V
20 – 5
= 3 A .The current exit from 20 V source is 12 A and
5
IA =
entering in 5 V source and 5 resistance is 3 A
Therefore, Power Delivered by 20V source is 20 × 12 = 240 W
Power Absorbed by 5V source is 5 × 3 = 15 W
Power Absorbed by 5 resister is I2R = 32 ×5 = 45 W
Power Absorbed by 3IA corrent source is 3IA V A
= 3 × 3 × 20 = 180 W
Thus power deliverd (240 W) = Power Absorbed (15 + 45 + 180) W
Ex.3For the circuit shown below the current I flowing through the circuit
will be
2V
(a)
1
2
A
(b)
1A
(c)
2A
(d)
4A
Sol.(c)The given circuit can be redrawn as
I
I
2V
2V
Clearly all the resistances are
I
connected in parallel
So, I = 2 A
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Ex.4 The current flowing through a capacitor in an AC circuit is :
(a)
Non-existent
(b)
Conduction current
(c) Displacement current
(d) None of the above
Sol.(c)We know total current density (J) is sum of displacement current
density & conduction current density i.e. JTotal = JD + JC
Also, Displacement current density is given as i.e.
J D  J  E and
PERSONAL REMARK : 
Ex.3 In the circuit shown, the
power supplied by the voltage
source is
1
1
1
Conduction current density is given as i.e. J C   E
±
In capacitor there is dielectric in between the two parallel plates.
So,  = 0.Therefore , JC = 0.For DC Supply  = 0 ,So, JD = 0
Therefore total current for DC supply = 0 A For AC although
, JC = 0 but as there is some frequency associated with AC i.e.
  0 .So there exist displacement current i.e. J D  0 .
Finally we can say that the current flowing through a capacitor
in an AC circuit is
Displacement current
1A
1
2A
1
(a) 0 W
1 I+3
I+2
1
(a) 4 W
(b) 2 W
(3) –2 W
Sol.(b)From given circuit, we observe that I1 =
or I = 0
Therefore power supplied by the
voltage source, P = VI
or P = 10 × 0 = 0 Watts
+
7
2
VA
2A
(a) 36 W
(b) 15 W
(c) 07 W
Now, applying KVL we get 36 – 15I –2 VA – 7 I = 0 ....(i)
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(d) 14 W
Sol.(d) Let the current I supplied by the 36 V source
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Ex.4 The total power consumed
in the circuit shown in the figure is
2VA
+
+
36 V
10V
10 – 2 (I + 3) – 2 (I + 2) = 0
current delivered by 2 V battery is = 1 A
Therefore, power delivered by 2 V battery is = V  I= 2  1
=2W
Ex.6 The power dissipated in the controlled source of the network shown
below is
15 
1A
+
–
1
(4) – 4 W

= 2A

I
1
3A
1

(b) 5 W
(c) 10 W (d) 100 W
Sol. Let the current I supplied by
the voltage source.Apply KVL in the
outer loop as shown
Ex.5 For the circuit given in figure below the power delivered by the 2 V
source is given by

10 V
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(c) 16 W (d)
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and VA = 7 I
....(ii)
or 36 – 15 I – 14 I – 7 I = 0
PERSONAL REMARK :

or 36 = 36 I or I = 1 A
Therefore, power dissipated in the controlled source is
2VA = 2  7 I  I = 14 W
Ex.7 The V – 1 relation for the networks shown in the given box is V = 4I
– 9. If now a resistor R = 2 is connected across it, then the value
of I will be.
[SSC-JE : 2011]
I
Network
R=2
V
(a) – 4.5 A
(b) – 1.5 A
(c) 1.5 A
(d) 4.5 A
Sol.(c) The given circuit
I
Network
R=2
V
From the given ,V – I relation ,V = 4I – 9
....(i)
from circuit ,V = IR
....(ii)
Put equation (i) in equation (ii) 4I – 9 = I × 2 or 2I = 9 or I = 4.5 A
Ex.8 The equivalent resistance between terminals X and Y of the network
shown is
[SSC-JE : 2012]
15 
5
Y
X
30 
10 
(a) 8 
(b) 



(c)



(d)



Sol.(c)The given circuit
5
15 
Y
X
10 
30 
Circuit is wheat stone bridge balance.
5
15 
X
Y Rxy =
10 
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Ex.9 The wave shape of current flowing through an inductor is
PERSONAL REMARK :
i

Sol.(b)The given circuit can be
2A
redrawn as
[SSC-JE : 2012]
0
t
T
2
The wave shape of voltage drop (v) across the inductor is
i
1A
u
0
2A
2
2V
u
u
(a)
1A
(b)
V
t
T
0
u
T
u
t
Power consumed
= (2)2 2 + 12 × 2 + 1 × 2
u
u
(c)
= 8 + 2 + 2 = 12 W
(d)
0
t
T
0
T
t
Sol.(d) The given wave shape
i
2A
0
T
t
di
therefore output is the derivative of input i.e.
dt
option (d) is correct choice.
We know that, V = L
Ex.10 A 10 µF and a 20 µF capacitor are in series. The combination is
supplied at 150 V from a sinusoidal voltage source. The voltage across
the 20 µF capacitor is then.
(a) 75 V
(b)
125 V
[SSC-JE : 2012]
(c) 100 V
(d) 50 V
Sol.(d) According to question
C1
C2
Ex.11The conditions at which the following potential divider is independent
of frequency, may be
(i)
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R 2 C2
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(ii)
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
R 2 C1
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(iii) R1 C1 << 1, R2 C2 << 1
(a) (ii) and (iv) are true
(c) (i) is true only
(iv) R1 + R2 +
1
1

C1 C 2
PERSONAL REMARK :

(b) (i) and (iii) are true
(d) (ii) is true only
Sol.(d) For indepndent of frequency time constants should be equal.So, R1
C1 = R2 C2 or
R1 C 2

R 2 C1
Ex.12 Two wires A and B of the same material but of different lengths L
and 2L have the radius r and 2r respectively. The ratio of specific
resistance will be
(a) 1 : 4
Sol.(c) R =
[SSC-JE : 2012]
(b) 1 : 8
(c) 1 : 1
(d) 1 : 2
l
l

A
r 2
R1 L1

R 2 L2
2
2
R 
L  2r 
R
2
  2  
  or 1 
2L  r 
R2 1
 R1 
Ex.13 A 20 micro farad capacitor is connected across an ideal voltage
source. The current in the capacitor
[SSC-JE : 2012]
(a) Will be very high at first, then exponentially decay and at steady
state will become zero.
(b) None of these are true.
(c) Will be zero at first, then exponentially rise
(d) Will be very high at first, then exponentially decay
Sol.(a) According to question
I
I
t
Current is high at first, then exponentially decay and at steady state
will become zero.
Ex.14
[SSC-JE : 2013]

i
+
e
V
–
The voltage (v) vs. current (i) curve of the circuit is shown below :
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v
PERSONAL REMARK :
100 V

88 V
0
i
0.5 A
Internal resistance of the source e is &
(a) 24 
(b) 4 
Sol.(d) Total resistance of circuit
RT 
(c) 10 
(d) 14 
100  88 12

 24 
0.5
0.5
Total resistances of circuit = 10 + source internal resistance
24 = 10 + r or r = 14 
Ex.15 Three inductors each of 60 mH are connect in delta. The value of
inductance of each arm of the equivalent star connection is &
(a) 10 mH
Sol.(a) Given that
(b) 15 mH
(c) 20 mH
(d) 30 mH
[SSC-JE : 2013]
L1 = 60 mH
L1
L2 = 60 mH
L2
L3 = 60 mH
L3
We know that
L1 
16.
Y is
L1L 2
60  60
3600


 20 mH
L1  L 2  L3 60  60  60
80
Three lamps are in circuit as shown in Figure, At what condition 100 W
lamp will have the maximum brightness ?
[SSC-JE : 2012]
40 W
K1 40 W
100 W
K2
K3
(a) K1 is closed, K2 is open and K3 is also open
(b) Both (c) and (d)
(c) Key K1 is closed, K2 is open and K3 is closed
(d) Key K1 is open, K2 is closed and K3 is open
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Sol.(d) The given circuit
PERSONAL REMARK :

40 W
K 1 40 W
100 W
K3
K2
Key K1 is open, K2 is closed and K3 is open.
Ex.17 For the network shown in fig.1, find the current in each resistor
using super position principle.
[SSC-JE : 2012]
10 
5
3
50 V
25 V
Figure 1
Ex.18 In the circuit, v is the input voltage applied across the capacitor of
2F. Current through the capacitor is &
[SSC-JE : 2013]
V
10
V +
–
3
2F
1 2
4
5
t
–12
20
10
(a)
0 1
–10
3
t (b)
5
0 1
–20
0 1
–11
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5
22
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11
10
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3
3
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3
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Sol.(d) The given figure
PERSONAL REMARK :

V
10
3 4
t
12
5
– 12
I
+
–
2F
First capacitor will be charged p to 10 V after
charging capacitor voltage VC 
iC
1
i.dt
C
dVc
 2  10  20 Amp
dt
It is discharging current of capacitor for positive cycle C = 20 Amp
 ve cycle current i  c
dv
 2  11  22
dt
Ex.19 (i) An oven operates on a 15.0 A current from a 120 V source. How
much energy will it consume in 3.0 h of operation? [SSC-JE : 2013]
How many 100 W light bulbs connected to a 120 V supply can be
turned on at the same time without blowing a 15.0 A fuse?
[SSC-JE : 2013]
(iii) 3.0 A, 125 V circuit contains a 10. W resistor. What resistance must
be added in series for the circuit to have a current of 5.0 A?
[SSC-JE : 2013]
Ex.20 In the following circuit, find the total resistance, R3, V2 and I4.
(ii)
R1
R3 = ?
R2
Vt
R5
It
R1 = 9 ,
R5 = 36 ,
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Vt = 12 V,
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R4 = 12 ,
It = 1.0 A
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21. Three resistor, each of ‘R’are connected in star. What is the value
of equivalent delta connected resistors ?
[SSC-JE : 2014]
R

(b) 2 R
2
Sol.(d) The given figure
(a)
(c)
R

3
PERSONAL REMARK :

(d) 3 R 
1
R1
RB
RC
R2
R3
3
2
RB
RA 
R1R 2  R 2 R 3  R 3R1
R1R 2  R 2 R 3  R 3R1
, RB 
R1
R2
Rc 
R 1R 2  R 2 R 3  R 3 R 1
R3
R  R  R  R  R  R 3R 2
RA 

 3R
R
R
22. Find R3 for the circuit shown in figure :
[SSC-JE : 2014]
R1
+
100 k
–
R2
R3
10 mA
50 mA
(a) 25 milli ohm
(c) 25 kilo ohm
Sol.(c) The given circuit
(b) 25 ohm
(d) 25 mega ohm
R1
+
100 k
–
R2
R3
10 mA
50 mA
Apply KCL 50 mA = 10 mA + I, I = 40 mA
By current division rule
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Current in R2 = 10 mA 
PERSONAL REMARK :
R3
 50 mA
R 3  100 k

10 R3 + 1000 = 50 R3
R3 
1000
 25 k
40
Ex.23 Two electric bulbs have tungsten filament of same thickness. If one
of them given 60 W and the other gives 100 W, then [SSC-JE : 2014]
(a) 60 W lamp filament has shorter length
(b) 100 W lamp filament has longer length
(c) 60 W lamp filament has longer length
(d) 60 W and 100 W lamp filaments have equal length
Sol.(c) R 
i.e.
 R 1  R 2
or
 
A
1 A  2
R 1 1
a linear relation

R2 2
So, the bulb with high resistance has more length of filament.
Ex.24 Two 100 W, 200 V lamps are connected in series across a 200 V
supply. The total power consumed by each lamp will be watts
(a) 25
(b) 50
(c) 100
(d)
200
Sol.(a)Resistance of each bulb
[SSC-JE : 2014]
V2
V 2 2002
P

 400 
, So R 
R
P
100
Equivalent circuit
I
400 
400 
100 W
100 W
+
–
200 V
Current in circuit I 
V 200

 0.25 A
R 800
Power consumed by each lamp is
I 2 R  0.252  400  25
Ex.25 The voltage across 5 – H inductor is
[SSC-JE : 2014]
 t 2 , t  0
V(t) = 
0,
t0
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Find the energy stored at t = 5 s. Assume zero initial current.
(a) 0.625 kJ
(b) 3.125 kJ (c) 156.25 kJ (d) 312.5 kJ
PERSONAL REMARK :

Ex.1. As per IE rules the permissible
30t 2 , t  0
Sol.(a) V(t)  
 0 , t0
variable variation of voltage at
the consumer end is :
1
Energy stored in inductor is given by E  LI 2 and current thorugh
2
1
inductor is i L   VL dt
L
Taking square of equation , i 2L 
[SSC-JE : 2014]
(a)  10%
(c)  2%
Sol.(d) + 6%
(b)  12%
(d)  6%
1
VL2 dt
L
5
S
 t3 
1
1
1
E  L *  30t 2 dt & E   30  
2
2
L0
 3 0
E
30  53  03 

2  3 
E
30
 125  5  125  625 joule = 0.625 kJ
2
Ex.26.The maximum demand of a consumer is 2 kW and his daily energy
consumption is 20 units. His load factor is :
(a) 10.15 %
(b) 41.6 %
Sol.(b) Load factor 

(c) 50 %
[SSC-JE : 2014]
(d) 21 %
Actual load
 100
Max. demand
20 / 24
 100  41.66 %
2
Ex.27A consumer has annual consumption of 7,00,800 units. If his maximum
demand is 200 kW. The load factor will be :
(a) 20%
(b)
40%
(c)
50%
[SSC-JE : 2014]
(d) 70%
Sol.(b) Annual consumption = 700800
Consumption per day 
700800
365
Consumption per hour 
Load factor 
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 80 kW / h
365  24
80
 100  40 %
200
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Ex.28 Two wires A and B have the same cross-section and are made of
PERSONAL REMARK :

the same material. RA = 800 and RB = 100 . The number of times
A is longer than B is :
(a) 6
(b)
Sol.(a) R  
[SSC-JE : 2014]
2
(c) 4
(d) 5
RA  RB

 
&
1
A 2
A
R A 1

RB 2
6000

1
or 100  
2
or 1 is 6 times  2
Ex.29 Determine the voltage at point C shown below with respect to
ground :
[SSC-JE : 2014]
A
100 
120 V
C
50 
B
(a) 120 V
(b) 40 V
(c) 70 V
(d) 80 V
Sol.(b) The given figure
100 
120 V
50 
By voltage divider rule
50
 120  40 Volt
150
Ex.30 Find the current through 5 resistor :
10 A
2
[SSC-JE : 2014]
5
(a) 7.15 A
(b) 5A
(c) 2.85 A
Sol.(c) By current divison rule current in 5  resistor is

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(d) 3.5 A
2
 10  2.857 A
5 2
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Ex.31
Total capacitance between the points L and M in figure is :
L
2µF 2µF 2µF
N
O
P
PERSONAL REMARK :

M
1µF 1µF
[SSC-JE : 2014]
1µF
(a) 1.45 µF
Sol.(c)
N
L(2 + 1)
(b) 1.85 µF
O
P
(c) 2.05 µF
(d) 4.05 µF
[{(2 1) || 2} 1]
{(2  1) || 2}  1 || 2  1  2.05
2+1
O
N
P
F
( 3|| 2) + 1
F
( 2.2|| 2) + 1
F
Ex.32A residential flat has the following average electrical consumptions
per day :
[SSC-JE : 2014]
(i)
4 tube lights of 40 watts working for 5 fours per day.
(ii)
2 filaments of 60 watts working for 8 hours per day.
(iii)
1 water heater rated 2 kW working for 1 hour per day.
(iv)
1 water pump of 0.5 kW rating working for 3 hours per day.
Calculate the cost of energy per month if 1 kWh of energy (i.e. 1
unit of energy costs. Rs. 3.50)
Sol. Try yourself
Ex.33fn, x, fp= esa 4  Áfrjks/k esa Áokfgr /kkjk dh x.kuk dhft,A
[UPRVNL : 2014]
V1
5A
(a) 3.4 A
Sol. Try yourself
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6V

(b) 1.3 A
V2
2A
(c) 2.8 A

(d) 4.6 A
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Ex.34 fn;s x;s fp= esa VA, VB, VC, VD, VE dk eku fudkfy,A VA fcanq
A vkSj Hkwfe (ground) ds chp dh oksYVrk gSA blh çdkj vU;
oksYVrk,a gSA
[UPRVNL : 2014]
A
PERSONAL REMARK :

B
3 K
+
C
18 V
5 K
E
1 K
D
(a) + 16 V, +10 V, + 16 V, 0 V,  2 V
(b) + 10 V, + 16 V, + 10 V, 0 V  2 V
(c) + 16 V, + 16 V, + 10 V, 0 V  2 V
(d) + 16 V, + 16 V + 10 V,  4 V,  2 V
Sol. Try yourself
Ex.35 fuEu fn”V/kkjh ifjiFk (dc circuit) esa (Branch) ‘kk[kk A  B esa
[UPRVNL : 2014]
/kkjk dh x.kuk dhft,A
16 A
A

5A

4A



B
7A
(a)
5.75 A
Sol. Try yourself
(b) 6.85 A
(c) 4 A
(d) 8.6 A
Ex.36 fuEu ifjiFk esa 6  Áfrjks/k esa fo|qr /kkjk fdruh gksxh\
10 V 

(a) 1.23 A
Sol. Try yourself
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
(b) 1.11 A
3A

[UPRVNL : 2014]
(c) 2.03 A
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(d) 0.31 A
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
37. fuEu ifjiFk esa fcnaq A vkSj B ds chp dk rqY; Áfrjks/k (equivalent
resistance) D;k gS\
[UPRVNL : 2014]

PERSONAL REMARK :


A
B
C
D


(a) 2.5 
(b) 4 
(c) 8.4 
(d) 6.8 
Sol. Try yourself
38. fuEu ifjiFk esa L=ksr ls fy;k tkus okyh dqy /kkjk (Total current)
,oa Áfrjks/k ds fljks ij foHkokUrj (potential difference) fdruk
gksxk\
[UPRVNL : 2014]
1.6 V 

I1
I2 
(a) 0.56 A, 0.8 V
(b) 0.4 A, 0.8 V
(c) 0.85 A, 0.6 V
(d) 0.9 A, 0.8 V
Sol. Try yourself
39. fuEu ifjiFk esa tky fo’ys”k.k (mesh analysis) }kjk I1 dk eku
[UPRVNL : 2014]
fudkfy,A

8 V +–
(a) 3.52 A
Sol. Try yourself
40.


3A
I1
5A
(b) 5.6 A
+
–
2V
(c) 6.3 A
Find the current through R1 in the given circuit :
(d) 1.71 A
[TTA : 2012]
R2 = 68
R1 = 120
(a) 0.16 A
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Is1 = 0.2A
(b) 0.24 A
Is2 = 0.04A
(c) 0.2 A
(d) 0.04 A
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Sol.a) Given circuit
PERSONAL REMARK :

A R2 = 68
R1 = 120
Is1 = 0.2A
Is2 = 0.04A
KCL at node A
Flowing current through R1
I = Is1  Is2 = 0.2  0.04 = 0.16 A
41.
A capacitor capable of storing 1 J of energy at 100 V dc supply.
The value of capacitance will be?
[TTA : 2013]
(a) 100 F
(b) 200 F
(c) 50 F
(d) 400 F
Sol.(c) We know that,
E
1
CV 2 or c
2
The resistance of a wire is R . If it is stretched to n times its original
length, the new resistance (in ) will be [TTA : 2013]
(a) R
(b) R/n
(c) n2R
(d) R/n2
Sol.(c)
42.
43.
Which has a higher resistance: a 2 KW electric heater or a 200 W
filament bulb, both marked for 230 V [TTA : 2013]
(a) 200 W Bulb
(b) 2 KW heater
(c) can’t say
(d) both are equal
Sol.(a) We know that P 
V2
1
or P 
R
R
So, 2 kW electric heater
less resistance and 200 W filament bulb has higher resistnce.
The resistance of a wire is 5 at 50 0 C and 6 at
1000 C. The resistance of the wire at 00 C will be - [TTA : 2013]
(a) 1 
(b) 4 
(c) 3 
(d) 2 
Sol. Try yourself
44.
45.
In the given circuit, R1 > R2, power dissipation will be [TTA : 2013]
R1
R1
V
(a) Greater in R1
(c) Equal in both
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(b) Greater in R2
(d) Can’t assess
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Sol.(b) The given circuit
PERSONAL REMARK :

R1
R1
V
In given circuit R1 and R2 in parallel So, R1 and R2 across voltage
will be same.
we know that P 
1
, therefore power dissipation will greater in
R
R 2.
The filament of 60 W and 100 W bulbs are of the same length then
(a) 60 filament is thicker
[TTA : 2013]
(b) 100 W filament if thicker
(c) both are of same thickness
(d) can’t be assessed
Sol.(b) 100 W bulb filament is thicker because 100 W bulb has less resistace.
46.
A student connects four cells each of emf 2V and internal resistance
0.5, in series but the one cell has its terminal reversaed. This sends
current in a 2 resistor. The current is
[TTA : 2013]
(a) zero
(b) 1A
(c) 1.5 A
(d) 2 A
Sol.(b) Given that four cells connected in series each of emf 2 V &
internal resistance 0.5 but one cell is connect opposite direction.
Total internal resistance = 0.5  + 0.5 + 0.5 + 0.5 = 2 
47.
A
R i = 2
2 V +–
2 V –+
+
2 V –
–
2V+
VT = 4 V
Therefore, I L 
48.
IL
 R L
VT
4
4

 1 A
Ri  RL 2  2 4
A 100 resistor is connected to 220 V - 50 HZ AC supply. What is
the net power consumed for a complete cycle [TTA : 2013]
(a) 242 W
(b) 484 W
(c) 220 W
(d) 100 W
Sol.(b) Given that R = 100 , AC supply = 220 V - 50 Hz
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PERSONAL REMARK :


220 V-50 Hz
We know that, P 
49.
V 2 (220) 2

 484 W
R
100
If the network of figure - I and T-netwrok of figure- II are equivalent,
then the values of R,. R2 and R3 will be respectively - [TTA : 2013]
R1
R2
16
R3
24
24
(a) 9, 6  & 6 
(c) 9, 6  & 9 
(b) 6 , 6 and 9 
(d) 6, 9 & 6 
Sol.(a) The given circuit
R1
R2
R3
16
24
24
We know that
R1 
R12  R13
24  24

9 
R12  R 23  R 31
64
So, R 2 
R 21  R 23
24  16

6
R12  R 23  R 31
64
R3 
50.
R 31  R 23
24  16

6
R12  R 23  R 31
64
In the circuit shown in the given figure, power dissipated in the 50 
resistor is
[TTA : 2013]

I

8A
(a) zero
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(b) 3.2 KW
10A
(c) 320 W
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CHAPTER- 1 : BASIC FUNDAMENTALS OF CIRCUITS (NETWORK)
Sol.(a) Given circuit
PERSONAL REMARK :


I

8A

10A
Equivalent circuit of the given circuit
140 
Vx
I
50 
8A
Vx Vx

 10  0
50 140
At node Vx 8 
51.
10 A
The maximum power that a 12 V d.c. source with an internal
resistance of 2 can supply to a resistive load is [TTA : 2013]
(a) 12 W
(b) 18 W
(c) 36 W
(d) 48 W
Sol.(c) Given that Vs = 12 V dc, Ri = 2 
R i= 2 
+
 12 V
RL
–
we know that P 
52.
V 2 122

 36 W
R
2
The current flowing through the voltage source in the below circuit
is
[TTA : 2013]
3V
(a) 1.0 A

(b) 0.75 A
0.25 A
(c) 0.5 A
(d) 0.25 A
Sol.(c)In the given circuit flowing current through i.e.
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Vs 
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V
 0.25   0.25  0.75  0.25  0.5 A
4
R
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PERSONAL REMARK :
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
43