CHAPTER 5 SERIES AND PARALLEL NETWORKS
EXERCISE 19, Page 47
1. The p.d.’s measured across three resistors connected in series are 5 V, 7 V and 10 V, and the
supply current is 2 A. Determine (a) the supply voltage, (b) the total circuit resistance and
(c) the values of the three resistors.
(a) Supply voltage, V = 5 + 7 + 10 = 22 V
(b) Total circuit resistance, R T 
V 22

= 11 
I
2
V1 5
 = 2.5 ,
I 2
V2 7
 = 3.5 
I
2
(c) R1 
R2 
and R 3 
V3 10

=5
I
2
2. For the circuit shown below, determine the value of V1. If the total circuit resistance is 36 
determine the supply current and the value of resistors R1, R2 and R3
Supply voltage, 18 = V1 + 5 + 3 from which, voltage, V1= 18 – 5 – 3 = 10 V
From Ohm’s law, supply current, I =
Resistance, R1 =
V 18

= 0.5 A
R T 36
V1 10

= 20 
I 0.5
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Resistance, R 2 =
V2
5

= 10 
I
0.5
Resistance, R 3 =
V3
3

=6
I 0.5
3. When the switch in the circuit shown is closed the reading on voltmeter 1 is 30 V and that on
voltmeter 2 is 10 V. Determine the reading on the ammeter and the value of resistor R X
Voltage across 5  resistor = V1  V2  30  10 = 20 V
Hence, current in 5  resistor, i.e. reading on the ammeter =
Total resistance, R T 
V5 
5

20
=4A
5
VT 30

 7.5  , hence R X = 7.5 – 5 = 2.5 
I
4
4. Calculate the value of voltage V in the diagram below.
5
 5 
By voltage division, voltage, V = 
  72   72 = 45 V
8
 53
5. Two resistors are connected in series across an 18 V supply and a current of 5 A flows. If one of
the resistors has a value of 2.4  determine (a) the value of the other resistor and (b) the p.d.
across the 2.4  resistor.
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The circuit is shown above.
(a) Total resistance, R T 
18
 3.6  , hence R X = 3.6 – 2.4 = 1.2 
5
(b) V1  5  2.4 = 12 V
6. An arc lamp takes 9.6 A at 55 V. It is operated from a 120 V supply. Find the value of the
stabilising resistor to be connected in series.
A circuit diagram is shown below.
The purpose of the stabilising resistor R S is to cause a volt drop VS – in this case equal to 120 – 55,
i.e. 65 V. Hence, R S 
VS 65

= 6.77 
I 9.6
7. An oven takes 15 A at 240 V. It is required to reduce the current to 12 A. Find (a) the resistor
which must be connected in series, and (b) the voltage across the resistor.
(a) If the oven takes 15 A at 240 V, then resistance of oven, R oven 
240
= 16 A
15
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A circuit diagram is shown above.
If the current is reduced to 12 A then the total resistance of the circuit, R T 
and
R T  R S  R oven
i.e.
20  R S  16 from which, series resistor, R S  20  16 = 4 
V 240

= 20 
I
12
(b) Voltage across series resistor, VS  I  R S  12  4 = 48 V
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EXERCISE 20, Page 53
1. Resistances of 4  and 12  are connected in parallel across a 9 V battery. Determine (a) the
equivalent circuit resistance, (b) the supply current, and (c) the current in each resistor.
(a) Equivalent circuit resistance, R T 
(b) Supply current, I =
(c) I1 
4 12 48

=3
4  12 16
(or use
1
1 1
  )
R T 4 12
V 9
 =3A
RT 3
9
9
 12 
= 2.25 A, I 2 
= 0.75 A (or, by current division, I1  
  3 = 2.25 A
12
4
 4  12 
 4 
and I 2  
  3 = 0.75 A)
 4  12 
2. For the circuit shown determine (a) the reading on the ammeter, and (b) the value of resistor R.
(a) V = 3  5 = 15 V. Hence, ammeter reading, I6  
(b) IR = 11.5 – 3 – 2.5 = 6 A hence, R =
V 15

= 2.5 A
6 6
V 15

= 2.5 
I
6
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3. Find the equivalent resistance when the following resistance’s are connected (a) in series (b) in
parallel: (i) 3  and 2  (ii) 20 k and 40 k (iii) 4 , 8  and 16 
(iv) 800 , 4 k and 1500 
(a)(i) Equivalent resistance, R T = 3  + 2  = 5 
(ii) Equivalent resistance, R T = 20 k + 40 k = 60 k
(iii) Equivalent resistance, R T = 4  + 8  + 16  = 28 
(iv) Equivalent resistance, R T = 800  + 4 k + 1500  = 800 + 4000 + 1500 = 6300  = 6.3 k
(b) (i)
6
1 1 1 5
  
from which, equivalent resistance, R T = = 1.2 
5
RT 3 2 6
(ii)
40
1
1
1
3



from which, equivalent resistance, R T =
= 13.33 k
3
R T 20 40 40
(iii)
16
1
1 1 1
7
   
from which, equivalent resistance, R T =
= 2.29 
7
R T 4 8 16 16
(iv)
1
1
1
1
13




R T 800 4000 1500 6000
from which, equivalent resistance, R T =
6000
= 461.54 
13
4. Find the total resistance between terminals A and B of the circuit shown below.
Total resistance between terminals A and B = 2 +
6  18
+ 1.5
6  18
= 2 + 4.5 + 1.5 = 8 Ω
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5. Find the equivalent resistance between terminals C and D of the circuit shown in Figure 5.30(b)
15 Ω in parallel with 15 Ω = 7.5 Ω
Three 15 Ω resistors in parallel = 5 Ω
Hence, total resistance between C and D = 15 + 7.5 + 5 = 27.5 
6. Resistors of 20 , 20  and 30  are connected in parallel. What resistance must be added in
series with the combination to obtain a total resistance of 10 . If the complete circuit expends
a power of 0.36 kW, find the total current flowing.
The circuit is shown below.
For the parallel branch,
1
1
1
1



R P 20 20 30
from which,
R P  7.5 
Hence, resistance to be added in series, R X  R T  R P  10  7.5 = 2.5 
Power, P = I2 R hence 0.36 103  I2 (10)
from which, total current flowing, I =
360
 36 = 6 A
10
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7. (a) Calculate the current flowing in the 30  resistor shown in the circuit below
(b) What additional value of resistance would have to be placed in parallel with the 20  and
30  resistors to change the supply current to 8 A, the supply voltage remaining constant.
(a) Total resistance, R T  4 
Hence, total current, I =
20  30
 4  12 = 16 
20  30
V 64

=4A
R T 16
 20 
and, by current division, I30   
  4  = 1.6 A
 20  30 
(b) If I = 8 A then new total resistance, R T2 
64
= 8  and the resistance of the parallel branch
8
will be: 8 – 4 = 4 
1 1
1
1

 
i.e.
where R X is the additional resistance to be placed in parallel
4 20 30 R X
from which,
1
1 1
1
 

R X 4 20 30
from which, R X = 6 
8. For the circuit shown, find (a) V1, (b) V2, without calculating the current flowing.
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5
 5 
By voltage division, V1 = 
  72   72 = 30 V
12
 57 
Hence, V2 = 72 – 30 = 42 V
or
7
 7 
V2 = 
  72   72 = 42 V
12
 57 
9. Determine the currents and voltages indicated in the circuit below.
1
1 1 1
  
from which, R P1 = 1 
R P1 2 3 6
R P2 
23
= 1.2 
23
Hence, total resistance, R T = 4 + 1 + 1.2 = 6.2 
I1 =
31
= 5 A, V1 = I1 (4)  5  4 = 20 V,
6.2
I2 =
V2 5
 = 2.5 A,
2 2
I3 
5
2
= 1 A,
3
3
V2 = 5 1 = 5 V and
I4 =
5
A,
6
I5 =
V3 = 5 1.2 = 6 V
V3 6
 =3A
2 2
and
I6 =
6
=2A
3
10. Find the current I in the circuit below.
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The circuit is reduced step by step as shown in diagrams (a) to (d) below.
(a)
(b)
(c)
(d)
From (d), IT 
24
=6A
4
 6 
From (b), I1  
  6  = 3.6 A
64
 5 
and from (a), I  
  3.6  = 1.8 A
 55
11. A resistor of 2.4  is connected in series with another of 3.2 . What resistance must be
placed across the one of 2.4  so that the total resistance of the circuit shall be 5 ?
The circuit is shown below.
2.4  in parallel with R X = 5 - 3.2 = 1.8 
i.e.
1
1
1


2.4 R X 1.8
from which,
1
1
1


 0.13888....
R X 1.8 2.4
Hence, resistor to be connected across the 2.4  resistor, R X =
1
= 7.2 
0.138888...
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12. A resistor of 8  is connected in parallel with one of 12  and the combination is connected in
series with one of 4 . A p.d. of 10 V is applied to the circuit. The 8  resistor is now placed
across the 4  resistor. Find the p.d. required to send the same current through the 8  resistor
The initial circuit is shown below.
Total resistance, R T = (8 in parallel with 12) + 4 =
Total current, I T 
8  12
 4  4.8  4 = 8.8 
8  12
10
 12 
= 1.1364 A and I8   
 1.1364  = 0.6818 A
8.8
 8  12 
When the 8  resistor is moved, the circuit is as shown below.
Voltage, V1  0.6818  8 = 5.4544 V
Current, I 4 
V1 5.4544

= 1.3636 A
4
4
Current, IT2  1.3636  0.6818 = 2.0454 A
Total resistance, R T2  12 
48
 12  2.6666... = 14.6666…
48
Hence, the p.d. required to send the same current through the 8  resistor,
V = IT2  R T2  2.0454 14.6666 … = 30 V
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EXERCISE 21, Page 56
1. For the circuit shown, AS is 3/5 of AB. Determine the voltage across the 120  load. Is this a
potentiometer or a rheostat circuit?
AS =
3
 400 = 240  hence SB = 400 – 240 = 160 
5
The simplified circuit is shown below:
For the parallel resistors, total resistance, R P 
160 120 160 120

= 68.57 
160  120
280
The equivalent circuit is now as shown below:
Total circuit resistance, R T  240  68.57  308.57 
Hence, supply current, IS 
200
 0.6482 A
308.57
Thus, volt drop, VSB = volt drop across the 120  load = IS  68.57 = 0.6482  68.57 = 44.44 V
The circuit is a potentiometer since the 400  shown in the original diagram has three connections
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2. For the circuit shown, calculate the current flowing in the 25  load and the voltage drop across
the load when (a) AS is half of AB, (b) point S coincides with point B.
Is this a potentiometer or a rheostat?
(a) When AS is half of AB, then total circuit resistance, R T  R AS  R L = 250 + 25 = 275 
Current flowing in load, I =
150
V
=
= 0.5454 A or 0.545 A
275
RT
Voltage drop across 25  load = I  25 = 0.5454  25 = 13.64 V
(b) When point S coincides with point B, R T  R AB  R L = 500 + 25 = 525 
Current flowing in load, I =
150
V
=
= 0.2857 A or 0.286 A
525
RT
Voltage drop across 25  load = I  25 = 0.2857  25 = 7.14 V
The circuit is a rheostat since the 500  shown in the original diagram has two connections.
3. For the circuit shown, calculate the voltage across the 600  load when point S splits AB in the
ratio 1:3.
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If AB is split in the ratio 1:3, then the number of parts is 1 + 3 = 4
Hence, 1 part =
1200
= 300  and 3 parts = 3  300  = 900 
4
The circuit simplifies to the circuit shown below:
For the parallel resistors, total resistance, R P 
900  600 900  600

= 360 
900  600
1500
The equivalent circuit is now as shown below:
Voltage across 360  = VSB = voltage across 600  of original circuit
=
(Alternatively, I T 
360
 250 = 136.4 V
300  360
250
= 0.379 
300  360
(by voltage division)
and VSB = IT  360 = 0.379  360 = 136.4 V)
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4. For the circuit shown, the slider S is set at halfway. Calculate the voltage drop across the 120 
load.
If the slider S is set at halfway, the circuit is as shown below.
120 Ω in parallel with 1000 Ω =
120 1000
 107.14 
120  1000
Hence the circuit is as below.
107.14


Voltage drop, V = 
 100 = 9.68 V = voltage across the 120 Ω load
 1000  107.14 
5. For the potentiometer circuit shown, AS is 60% of AB. Calculate the voltage across the 70 
load.
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AS = 60% of 150 Ω = 90 Ω. Hence, SB = 150 – 90 = 60 Ω
The circuit is thus shown below.
60 Ω in parallel with 70 Ω =
60  70
 32.31
60  70
The equivalent circuit is thus shown below.
 32.31 
Voltage drop, V = 
  240 = 63.40 V = voltage across the 70 Ω load
 90  32.31 
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EXERCISE 22, Page 58
1. For the circuit below, calculate (a) the absolute voltage at points A, B and C, (b) the voltage at A
relative to B and C, and (c) the voltage at D relative to B and A.
Currents are shown in the circuit diagram below.
Total resistance, R T  15  6 
Total current, I T 
5  20
 21  4 = 25 
5  20
100
=4A
25
5


Current, I1  
  4   0.8 A
 5  7  13 
(a) VEarth  0 V
and
 7  13 
I2  
  4  = 3.2 A
 7  13  5 
VD  4  6 = + 24 V = VC
VA = VD   I2  5  24  (3.2)(5) = + 40 V
VB = VC   I1  7   24   0.8 7  = + 29.6 V
(or VB = VA   I1 13  40   0.813 = + 29.6 V)
(b) VAB = VA  VB = 40 – 29.6 = 10.4 V
VAC = VA  VC = 40 – 24 = 16 V
(c) VDB = VD  VB = 24 – 29.6 = - 5.6 V
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VDA = VD  VA = 24 – 40 = - 16 V
2. For the circuit shown below, calculate (a) the voltage drop across the 7  resistor, (b) the current
through the 30  resistor, (c) the power developed in the 8  resistor, (d) the voltage at point X
w.r.t. earth, and (e) the absolute voltage at point X.
Currents are shown in the circuit diagram below.
Total resistance, R T  18 
Total circuit current, IT 
20  30
 18  12 = 30 
20  30
12
= 0.4 A
30
 30 
Current, I1  
  0.4  = 0.24 A
 30  20 
and
 20 
I2  
  0.4  = 0.16 A
 20  30 
(a) Voltage drop across the 7  resistor = I1  7  0.24  7 = 1.68 V
(b) Current through the 30  resistor = I 2 = 0.16 A
(c) Power developed in the 8  resistor = I12 (8)   0.24  8  = 460.8 mW
2
(d) VEarth  0 V
Voltage at point X, VX   I1  7  5   0.24 12  = + 2.88 V
(e) Absolute voltage at X means ‘the voltage at X with respect to earth’ = + 2.88 V
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3. In the bridge circuit shown below, calculate (a) the absolute voltages of A and B, and (b) the
voltage at A relative to B.
Currents are shown in the circuit diagram below.
Total resistance, R T 
3000  24
= 23.81 
3000  24
and
total current, IT 
30
 1.26 A
23.81
24


 3000 
Current, I1  
 1.26  = 0.01 A and current, I 2  
 1.26  = 1.25 A
 3000  24 
 3000  24 
More simply, from the circuit, I1 
30
30
= 0.01 A and I 2 
= 1.25 a
3000
24
(a) VA =  I1 1k  0.011000 = 10 V
VB =  I2 8  1.25 8 = 10 V
(b) VAB = VA  VB = 10 – 10 = 0 V
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EXERCISE 23, Page 60
1. If four identical lamps are connected in parallel and the combined resistance is 100 , find the
resistance of one lamp.
If each lamp has a resistance of R then:
1
1 1 1 1 4
    
100 R R R R R
and
R = 4  100 = 400  = resistance of a lamp
2. Three identical filament lamps are connected (a) in series, (b) in parallel across a 210 V supply.
State for each connection the p.d. across each lamp.
(a) In series, p.d. across each lamp =
210
= 70 V
3
(b) In parallel, p.d. across each lamp = 210 V
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