CLUSTER LEVEL WORK SHOP
SUBJECT
PHYSICS
QUESTION BANK (ALTERNATING CURRENT ) DATE: 20/08/2016
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What is the phase difference between the voltage across the inductance and capacitor in series AC circuit?
Ans. Phase difference 1800
In a series of LCR circuit VL = VC ≠ VR , What is the value of power factor for this circuit?
Ans: power factor = 1
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The instantaneous current and voltage of an a.c. circuit are given by
i = 10 sin 300 t A and
V = 200 sin 300 t V.
What is the power dissipation in the circuit?
Ans:
It is given that:
i = 10 sin 300 t A
V = 200 sin 300 t V
∴ i0 = 10 A and V0 = 200 V
∴ Average power dissipation = V0i0
= 200 × 10 = 2000 W
In a series LCR circuit the voltage across an inductor, a capacitor and a resistor are 20V, 20V and 60V
respectively. What is the phase difference between the applied voltage and the current in the circuit?
Ans:
VL= VC ,(XL = XC) than circuit is said to be in resonance
Z = R ,Therefore phase difference V And I is 00
What is the average value of the emf for the shaded part of graph?
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Ans: average value of the emf for the shaded part = 0.636 V0 = 0.636 ×314 = 200V
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The instantaneous current from an a.c. source is I = 5 sin (314 t) ampere. What are the average and 1
rms values of the current?
ANS:
average value of current = 0, rms value of current I 0/ 2 =
5
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A 0.707 ×5 = 3.53 A
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What is the inductive reactance of a coil if current through it is 800 mA and the voltage across it is 40 1
V?
ANS:
XL= V/I = 40/ 800 m A = 40000/800 = 50 ohm
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A transformer has 300 primary turns and 2400 secondary turns .If the primary supply voltage is 230 V, 1
what is the secondary voltage?
Ans:
NP/NS = VP/ VS ; = 230 ×2400/300 = 1.84 kV
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A.C. power is transmitted from one station to another at highest possible voltage. Why?
ANS: As the voltage of A.C. power transmitted from one station to another is very large, the magnitude of the
current is very low. Therefore, the power loss is reduced in transmission of power as the power loss is given by
P = I 2R
10 What is wattless current?
In the purely inductive and capacitive circuit power loss is zero in such a circuit current fallowing is called
wattles current
11 Why is the use of A.C. voltage preferred over D.C.voltage? Give two reasons.
Solution:
The use of A.C. voltage is preferred over the use of D.C. voltage because of the following
reasons:
i) The loss energy in transmitting the A.C. voltage over long distances with the help of step
up transformers is negligible as compared to D.C. voltage.
ii) A.C. voltage can be stepped up and stepped down as per the requirement using a
transformer.
12 What is average value of AC voltage represented by V = 220 sin314t volts over the time interval
a. 0 to π/ω
b. 0 to 2π/ω
Ans: we know that T = 2π/ω
Here given t = π/ω = T/2 i.e. half period, Vav = 2 Vo / π = 7/11 Vo = 0.636 ×220 = 140 V
Over a complete cycle, Vav = 0.
13 Calculate power factor of an AC circuits having a resistance R and Inductance L connected in series.
Given angular frequency of source is ω.
Ans:
𝑍 = 𝑅 2 + 𝑋𝐿2 = 𝑅 2 + 𝜔 2 𝐿2
𝑅
𝑅
Power factor = cos(ᶲ) = 𝑧 = 2 2 2
1
1
2
2
2
𝑅 +𝜔 𝐿
14 In the figure below if the frequency of the A.C mains is increased. How will the current be affected?
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R
A.C Mains
Ans: R is not affected by frequency. Hence no change in current.
15 A bulb B and a capacitor C are connected in series with a.c. mains. The bulb glows with some
brightness. How will the glow of the bulb will change when a mica sheet is introduced between
the plates of capacitor.
C
B
AC
A
C
By introducing mica sheet between the plates of capacitor,
the capacitance increases.
m Glow of bulb decreases
Hence Capacitive reactance XC =1/ ω C ,decreases.
ai
n
s
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16 A bulb B and an inductor L are connected in series with a.c mains through key. The switch is
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closed and after some time an iron rod is inserted into the interior of the inductor. How will the
glow of the bulb will change?
L
B
AC
A
ANS: XL = ωL, Inductance increases. Hence inductive reactance increases.
Low amount of voltage appears across the bulb. Hence glow of bulb decreases.
17 Prove that an ideal capacitor in an AC does not dissipated power.
Ans:
P = Vrms × I rms cos , for ideal capacitor R = 0, therefore cos  = 0, P = 0
18 In a series LCR circuit the value of inductance is kept fixed at resonance but the resistance is doubled.
How will it affect the sharpness of resonance?
Ans:
Q = ωr L/ R;
Q’ = ωr L/ R’ = Q’ = ωr L/ 2R
Q’ = Q/ 2 , half
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21 An electric bulb B and a parallel plate capacitor C are connected in series to the a.c.mains as
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shown in the given figure. The bulb glows with some brightness.
How will the glow of the bulb be affected on introducing a dielectric slab between the plates
of the capacitor? Give reasons in support of your answer.
Solution:
The bulb will glow brighter.
Reasons:
The impedance of a capacitor is , X C =
without dielectric.
If a dielectric is introduced inside a capacitor, then the new capacitance will be C’ = KC and
the new impedance will be X C’ =
.
Therefore, the impedance has decreased.
This will result in higher current through the circuit and the bulb will glow brighter.
22 In a series LCR circuit, define the quality factor (Q) at resonance. Illustrate its significance by
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giving one example.
Show that power dissipated at resonance in LCR circuit is the maximum.
Solution:
The quality factor (Q) is defined as:
Where,
→ Resonant frequency
L → Inductance
R → Resistance
Another expression for quality factor is
Where,
2
is the band width.
So, larger the value of Q, the smaller the band width and sharper is the resonance.
Power =
Hence, power =
∴ Power dissipated will be the maximum at resonance
23 A resistor of 200 Ω and a capacitor of 40 μF are connected in series to 220 V a.c. source
with angular frequency (ω) = 300 Hz. Calculate the voltages (rms) across the resistor and the
capacitor. Why is the algebraic sum of these voltages more than the source voltage? How do
you resolve this paradox?
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Voltage drop across resistor = IR
= 1.02 × 200
= 204 V
Voltage drop across capacitor = IXC
Total voltage drop
= 204 + 85
= 289 V
Kirchoff’s voltage law will be valid only if we consider instantaneous values of voltage. 85 V
and 204 V are rms values of voltage and hence, when they are added there value becomes
more than the source voltage
24 A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate
(i) The resistance of the bulb;
(ii) The rms current through the bulb
Ans:
P = 100 W Vrms = 220 V
(a) Resistance, R
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(b) P = Irms Vrms
25 An alternative voltage given by V = 140 sin 314t is connected across a pure resistor of 50 Ω.
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Find
(i) The frequency of the source.
(ii) The rms current through the resistor.
Ans:
Given V = 140 sin 314t, R = 50Ω
(i) Comparing with V = V0 sin t
Thus, V0 = 140 V
ω = 314
2π = 314
26 A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the
expression for the impedance of the circuit. Plot a graph to show the variation of current with
frequency of the source, explaining the nature of its variation.
Solution:
Let an alternating Emf is E = E0 sint is applied to a series combination of inductor L,
capacitor C and resistance R. Since all three of them are connected in series the current
through them is same. But the voltage across each element has a different phase relation
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with current.
The potential difference VL, VC and VR across L, C and R at any instant is given by
VL = IXL, VC = IXC and VR = IR
Where I is the current at that instant.
XL is inductive reactance and
XC is capacitive reactance.
VR is in phase with I. VL leads I by 90° and VC lags behind I by 90°
In the phases diagram,
VL and VC are opposite to each other. If VL > VC then resultant (VL − VC) is represent by OD.
OR represent the resultant of VR and (VL − VC). It is equal to the applied Emf E.
The term
is called impedance Z of the LCR circuit.
Emf leads current by a phase angle 
When resonance takes place
Impedance of circuit becomes equal to R. Current becomes maximum and is equal to
This is the condition for resonance.
When at resonance f = f0 the current in the circuit is maximum and hence impedance of the
circuit is maximum for values of f less than or greater than f0 comparatively small current
flames in the circuit.
27 a) For a given a.c., i = im sin t, show that the average power dissipated in a resistor R over a
complete cycle is
R.
(b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the
bulb.
Solution:
(a)
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(b)
Power of the bulb, P = 100 W and Voltage, V = 220 V
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a.c source generating a voltage
is connected to a capacitor of An
capacitance C. Find the expression for the current i, flowing through it, plot a graph
of v and i versus ωt to show that the current is , ahead of the voltage.
A resistor of 200 Ω and a capacitor of 15 μF are connected in series to a 220 V, 50 Hz a.c
source. Calculate the current in the circuit and the rms voltage across the resistor and the
capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes,
resolve the paradox.
Ans:
A.C source containing capacitor:
Alternating emf supplied is:
Potential difference across the plates of capacitor
At every instant, the potential difference V must be equal to the emf applied i.e.,
Or, q = C
It I is instantaneous value of current in the circuit at instant t, then
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Numerical:
Here, r = 200 Ω, C = 15 μF = 15 × 10−6 F
E = 220 V, f = 50 Hz, I =?
This is because these voltages are not in same phase and they cannot be added like
ordinary numbers.
32 Explain briefly, with the help of a labelled diagram, the basic principle of the working of an a.c
generator.
In an a.c generator, coil of N turns and area A is rotated at V revolutions per second in a
uniform magnetic field B.
Write the expression for the emf produced. A 100-turn coil of area 0.1 m2 rotates at half a
revolution per second. It is placed in a magnetic field 0.01 T perpendicular to the axis of
rotation of the coil. Calculate the maximum voltage generated in the coil.
Ans:
A.C generator principle:
Whenever a closed coil is rotated in a uniform magnetic field about an axis perpendicular to
the field, the magnetic flux linked with coil changes and an induced emf is set up across its
ends.
The construction of an ac generator is shown in the figure. Initially, the coil ABCD is
horizontal. The coil starts rotating clockwise and the arm AB moves up whileCD moves
down. By Fleming’s right hand rule, the induced current flows along ABCD.
In second half rotation, the arm CD moves up and AB moves down. The induced current
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flows in the opposite direction that is along DCBA. Thus, an alternating current flows in the
circuit.
The magnetic flux linked with the coil at any instant is
Φ = NBA cos ωt
Induced emf will be
Or E = E0 sin ωt, where E0 = NBA ω = Peak value of induced emf
Numerical:
N = 100, A = 0.1 m2, B = 0.01T
∴ Maximum voltage,
33 (a) Derive an expression for the average power consumed in a series LCR circuit connected
to a.c. source in which the phase difference between the voltage and the current in the circuit
is Φ.
(b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in
receiving circuits? Name the factors on which it depends.
Ans:
(a) Power in ac circuit
Voltage v in an ac circuit is:
which drives through the circuit a current i
i = im sin (ωt +Φ), where
and
Power
Calculating the average power, it is observed that the average of the term
cos (2ωt + Φ) is equal to zero.
Thus,
Average power,
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(b) The ratio
is called the quality factor or Q-factor.
The quality factor has high value in receiving circuits in order to get a sharp gain for the
desired channel frequency.
The quality factor depends on the values of the following:
i.
Inductance
ii.
Resistance
iii.
Capacitance
34 (a) Derive the relationship between the peak and the rms value of current in an a.c. circuit.
(b) Describe briefly, with the help of labelled diagram, working of a step-up transformer.
A step-up transformer converts a low voltage into high voltage. Does it not violate the
principle of conservation of energy? Explain.
Ans:
OR
(a) The instantaneous power dissipated in the resistor is
The average value of p over a cycle is:
are constants. Therefore,
By trigonometric identity,
The average value of cos 2 ωt is zero.
We have:
Thus,
The rms value in the ac power is expressed in the same form as dc power root mean square
or effective current and is denoted by Irms.
Peak current is
Therefore,
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(b)
In a transformer with Ns secondary turns and Npprimary turns, induced emf or voltage Es is:
Back emf = Ep =
EP = VP
Es = Vs
Thus, Vs =
… (i)
Dividing equations (i) and (ii), we obtain
If the transformer is 100% efficient, then
Thus, combining the above equations,
If Ns > Np, then the transformer is said to be step-up transformer because the voltage is
stepped up in the secondary coil.
No, the transformer does not violate the principal of conservation of energies. This can be
easily observed by the following equation:
Power consumed in both the coils is the same as even if the voltage increases or current
increases, their product at any instant remains the same.
35 A series LCR circuit is connected to a source having voltage v = vm sin ωt. Derive the
expression for the instantaneous current I and its phase relationship to the applied voltage.
Obtain the condition for resonance to occur. Define ‘power factor’. State the conditions under
which it is (i) maximum and (ii) minimum.
Ans:
v = vm sin ωt
Let the current in the circuit be led the applied voltage by an angleΦ.
The Kirchhoff’s voltage law gives
It is given that v = vm sin ωt (applied voltage)
.
On solving the equation, we obtain
On substituting these values in equation (1), we obtain
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Let
This gives
and
On substituting this in equation (2), we obtain
On comparing the two sides, we obtain
Or
And
The condition for resonance to occur
For resonance to occur, the value of im has to be the maximum.
The value of im will be the maximum when
Power factor = cos Φ
Where,
(i) Conditions for maximum power factor (i.e., cos Φ = 1)
i.
XC = XL
Or
ii.
R=0
(ii) Conditions for minimum power factor
iii.
When the circuit is purely inductive
iv.
When the circuit is purely capacitive