In conclusion, there are two requirements which must be
met in order to establish an electric circuit. The
requirements are:
1.There must be an energy supply capable doing
work on charge to move it from a low energy location
to a high energy location and thus creating an electric
potential difference across the two ends of the
external circuit.
2.There must be a closed conducting loop in the
external circuit which stretches from the high
potential, positive terminal to the low potential
negative terminal.
RESISTORS IN SERIES
- In a series circuit, the current is the same
at all points along the wire.
IT = I1 = I2 = I3
- An equivalent resistance is the resistance of a single
resistor that could replace all the resistors in a circuit.
The single resistor would have the same current through
it as the resistors it replaced.
RE = R1 + R 2 + R3
- In a series circuit, the sum of the voltage drops equal
the voltage drop across the entire circuit.
VT = V 1 + V 2 + V3
RESISTORS IN SERIES
12.4 Two resistances of 2 Ω and 4 Ω respectively are connected in
series. If the source of emf maintains a constant potential
difference of 12 V.
a. Draw a schematic diagram with an ammeter and a voltmeter.
12.4 Two resistances of 2 Ω and 4 Ω respectively are connected in
series. If the source of emf maintains a constant potential
difference of 12 V.
b. What is the current delivered to the external circuit?
Re = R1 + R2
=2+4
=6Ω
V
12
IT 

=2A
Re
6
c. What is the potential drop across each resistor?
V1 = I R 1
= 2(2)
=4V
V2 = I R 2
= 2(4)
=8V
PARALLEL CIRCUITS
- In a parallel circuit, each resistor provides a new path
for electrons to flow. The total current is the sum of the
currents through each resistor.
IT = I1 + I2 + I3
- The equivalent resistance of a parallel circuit
decreases as each new resistor is added.
1
1
1
1



RE R1 R2 R3
- The voltage drop across each branch is equal to the
voltage of the source.
V T = V1 = V2 = V 3
RESISTORS IN PARALLEL
KIRCHHOFF’S LAWS
An electrical network is a complex circuit consisting of
current loops. Kirchhoff developed a method to solve
this problems using two laws.
Law 1. The sum of the currents entering a junction is
equal to the sum of the currents leaving that junction.
I in  I out
This law is a statement of charge conservation.
A junction (j) refers to any point in the circuit where two
or three wires come together.
j
KIRCHHOFF’S LAWS
Law 2. The sum of the emfs around any closed current
loop is equal to the sum of all the IR drops around that
loop. (Ohm’s Law: V = IR)
  IR
This law is a statement of energy conservation.
Gustav Robert Kirchhoff
(1824-1887)
12.5 The total applied voltage to the circuit in the figure is 12 V and
the resistances R1, R2 and R3 are 4, 3 and 6 Ω respectively.
a. Determine the equivalent resistance of the circuit.
R2 and R3 are in parallel (RP)
1 1 1
 
Rp 6 3
R p= 2 Ω
RP and R1 are in series
Re = 4 + 2
=6Ω
b. What is the current through each resistor?
V 12
IT 
=2A

Re 6
I1 = 2 A (series)
V1 = I1R1
= 2(4)
=8V
The voltage across the parallel combination is therefore:
12 - 8 = 4 V each
4
V
I2 
 = 1.33 A
R2
3
4
V
 = 0.67 A
I3 
R3 6
12.6 Find the equivalent resistance of the circuit shown.
1 and 2 are in series:
1+ 2 = 3 Ω
this combination is in
parallel with 6:
1 1 1
 
Rp 3 6
RP = 2 Ω
this combination is in series with 3:
2+3=5Ω
this combination is in parallel with 4:
1 1 1
 
Rp 5 4
RP = 2.22 Ω = Req
12.7 A potential difference of 12 V is applied to the circuit in the
figure below. a. Find the current through the entire circuit
R1 and R2 are in parallel:
1 1 1
 
Rp 4 4
RP = 2 Ω
This combination is in series with R3:
2+4= 6Ω
This combination is now in parallel with R4:
1 1 1
  = 3 Ω = Req
Rp 6 6
V
12
IT 
=4A

Req
3
b. Find the current through each resistor.
IT = I3 + I4
V 12
I4 

R4 6 = 2 A
I3 = IT - I4
=4-2
=2A
The voltage for the parallel
combination is:
V' = V - I3 R3
= 12 - (2)(4)
=4V
V' 4
I1 
 =1A
R1 4
I2 = 2 - 1 = 1 A
EMF AND TERMINAL POTENTIAL DIFFERENCE
Every source of emf (Є) has an inherent resistance
called internal resistance represented by the symbol r.
This resistance is a small resistance in series with the
source of emf. The actual terminal voltage VT across a
source of emf with an internal resistance is given by:
VT = Є - I r
Units: Volts (V)
12.8 A load resistance of 8 Ω is connected to a battery whose
internal resistance is 0.2 Ω
a. If the emf of the battery is 12 V, what current is delivered to the
load?
Є = 12 V
RL = 8 Ω
r = 0.2 Ω
12
V  

I
= 1.46 A
R RL  r 8  0.2
b. What is the terminal voltage of the battery?
VT = Є - I r
= 12 - 1.46(0.2)
= 11.7 V
12.9 a. Determine the total current delivered by the source of emf
to the circuit in the figure. V = 24 V. The resistances are 6, 3, 1, 2
and 0.4 Ω respectively.
R1 and R2 are in parallel:
1 1 1
 
Rp 6 3
this combination is in series with R3:
2+1= 3Ω
this combination is now in parallel with R4:
1 1 1
 
Rp 2 3
RP = 1.2 Ω
RP = 2 Ω
finally the internal resistance r is in series giving the
equivalent resistance:
Req = 1.2 + 0.4
= 1.6 Ω
V
24
IT 

Req 1.6
= 15 A
b. What is the current through each resistor?
VT = Є - I r
= 24 - 15(0.4)
= 18 V
V
18
I4 
=9A

R4
2
V4 = VT = 18 V
I3 = IT - I4
= 15 - 9
=6A
V3 = I3R3
= 6(1)
=6V
V1 = V2 = 18 - 6 = 12 V each
V 12

I1 
=2A
6
R1
12
V
=4A

I2 
3
R2
Ammeters and Voltmeters
A
+
V
Voltmeter
Emf
Rheostat
-
Source of
EMF
Ammeter
Rheostat
Galvanometer
The galvanometer uses
torque created by small
currents as a means to
indicate electric current.
A current Ig causes the
needle to deflect left or
right. Its resistance is Rg.
20
10 0 10
N
The sensitivity is determined by the current
required for deflection.
S
20
Operation of an Ammeter
The galvanometer is often the working element of
both ammeters and voltmeters.
Rg
Ig
A shunt resistance in parallel
with the galvanometer allows
Rs
I
most of the current I to bypass
the meter. The whole device
Is
must be connected in series
with the main circuit.
I = Is + Ig
The current Ig is negligible and only enough to
operate the galvanometer. [ Is >> Ig ]
Operation of an Voltmeter
The voltmeter must be connected in parallel and
must have high resistance so as not to disturb the
main circuit.
A multiplier resistance Rm is
added in series with the
galvanometer so that very little
current is drawn from the main I
circuit.
Ig
Rg
Rm
VB
The voltage rule gives:
VB = IgRg + IgRm
CAPACITORS IN SERIES AND PARALLEL
These are the symbols used in different
arrangements of capacitors:
CAPACITORS IN SERIES
CAPACITORS IN SERIES
 Series capacitors always have the same charge.
 The voltage across the equivalent capacitor Ceq is the sum
of the voltage across both capacitors.
1
1
1
1



 ...
Ceq C1 C2 C3
Q T  Q1  Q 2  Q 3  ....
V  V1  V2  V3  ....
CAPACITORS IN PARALLEL
CAPACITORS IN PARALLEL
Parallel Capacitors always have
the same voltage drop across
each of them.
The charge on the equivalent
capacitor Ceq is the sum of the
charges on both capacitors.
Ceq  C1  C2  C3  ....
Q T  Q1  Q 2  Q 3  ....
V  V1  V2  V3  ....
19.9 a. Find the equivalent capacitance of the circuit.
C2 = 2 μF, C3 = 3 μF, C4 = 4 μF
V = 120 V
C2 and C4 are in series
C2 ,4
C2 C4
2
(
4
)


= 1.33 μF
C2  C4 2  4
C3 is now in parallel with C2,4
Ceq= C3 + C2,4
= 3 +1.33
= 4.33 μF
1
1
1


C2 ,4 C2 C4
b. Determine the charge on each capacitor.
The total charge of the system
QT = CeqV
= 4.33 (120)
= 520 μC
Q3 = C 3V
= 3(120)
= 360 μC
Q2 and Q4 have the same charge since they are in series:
Q2 = Q4 = QT - Q3
= 520 - 360
= 160 μC
Q3 = 360 μC, Q2 = Q4 =160 μC
c. What is the voltage across the 4 μF capacitor?
Q4 160
V4 

C4
4
= 40 V
The remaining voltage (120 - 40 = 80 V) goes through the
C2 capacitor.
RC CIRCUITS
A resistance-capacitance (RC) circuit is simply a circuit
containing a battery, a resistor, and a capacitor in series with
one another. An RC circuit can store charge, and release it at
a later time. A couple of rules dealing with capacitors in an
RC circuit:
1. An empty capacitor does not resist the flow of current, and
thus acts like a wire.
2. A capacitor that is full of charge will not allow current to
flow, and thus acts like a broken wire.
RC Circuits
When the switch is closed, the capacitor will begin
to charge.
RC Circuits
If an isolated charged capacitor is connected across a
resistor, it discharges:
19.10 Three identical resistors, each with resistance R, and a
capacitor of 1.0 x 10-9 F are connected to a 30 V battery with
negligible internal resistance, as shown in the circuit diagram
above. Switches SI and S2 are initially closed, and switch S3
is initially open. A voltmeter is connected as shown.
a. Determine the reading
on the voltmeter.
b. Switches Sl and S2 are now opened, and then switch S3 is
closed. Determine the charge Q on the capacitor after S3 has
been closed for a very long time.
After the capacitor is fully charged, switches S1 and S2
remain open, switch S3 remains closed, the plates are held
fixed, and a conducting copper block is inserted midway
between the plates, as shown below. The plates of the
capacitor are separated by a distance of 1.0 mm, and the
copper block has a thickness of 0.5 mm.
c. i. What is the potential
difference between the
plates?
ii. What is the electric field
inside the copper block?
iii. On the diagram, draw
arrows to clearly indicate
the direction of the electric
field between the plates.
iv. Determine the magnitude
of the electric field in each of
the spaces between the
plates and the copper block.