CHAPTER 37 KIRCHHOFF’S LAWS
EXERCISE 174, Page 392
1. Find currents I 3 , I 4 and I 6 in the circuit below.
By Kirchhoff’s law, I1  I 2  I3 i.e. 4 = 2 + I3 from which, I 3 = 4 – 2 = 2 A
Also, I3  I 4  I5
i.e.
I3  I 4  I5
i.e. 2 + I 4 = 1
And I5  I6  I1
i.e.
1  I6  4
from which, I 6 = 4 – 1 = 3 A
from which, I 4 = 1 – 2 = - 1 A
2. For the networks shown below, find the values of the currents marked.
(a) 3 + I 2 = 2
from which, I 2 = 2 – 3 = - 1 A
Also, 10 = 7 + I1  I 2
And 7 + I1 + 2 = I3
i.e.
10 = 7 + I1  (1)
i.e. 7 + 4 + 2 = I3
i.e.
from which, I1 = 10 + 1 – 7 = 4 A
I 3 = 13 A
400
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i.e. I 2 = 60 A
(b) 50 + 10 = I 2
Also, I1  I 2 = 100 from which, I 2  100  I1  100  60
And 100 + 20 = I3
And I3 = 20 + I 4
And I 4 + I5 = 20
i.e.
I 3 = 120 A
i.e. I 4 = I3 - 20 = 120 – 20
i.e.
i.e. I 2 = 40 A
i.e.
I 4 = 100 A
100 + I5 = 20 from which, I 5 = 20 – 100 = - 80 A
3. Use Kirchhoff's laws to find the current flowing in the 6  resistor of the circuit below and the
power dissipated in the 4  resistor.
The currents are labelled as shown in the diagram below.
Kirchhoff's voltage law is now applied to each loop in turn:
For loop 1:
40 = 5I 1 + 4I 2
(1)
For loop 2:
0 = 4I 2 - 6(I 1 - I 2 )
(2)
Equation (2) simplifies to:
0 = - 6I 1 + 10I 2
(3)
5  equation (1) gives:
200 = 25I 1 + 20I 2
(4)
401
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2  equation (3) gives:
0 = - 12I 1 + 20I 2
(5)
Equation (4) - equation (5) gives:
200 = (25I 1 - -12I 1 )
i.e.
200 = 37I 1
Hence, current, I1 =
200
= 5.4054 A
37
Substituting I 1 = 5.405 into equation (1) gives:
40 = 5(5.405) + 4I 2
40 = 27.025 + 4I 2
and
from which,
40 – 27.025 = 4I 1
I2 =
40  27.025 12.975
=
= 3.2438 A
4
4
Hence, the current flowing in the 6 Ω resistance is
i.e.
I 1 - I 2 = (5.4054 – 3.2438) = 2.162 A
Power dissipated in the 4  resistor =  I 2  R   3.2438   4 = 42.09 W
2
2
4. Find the current flowing in the 3  resistor for the network shown in circuit (a) below. Find also
the p.d. across the 10  and 2  resistors.
The currents are labelled as shown in the diagram below.
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Loop 1:
20 = 3I 1 + (6 + 10)I 2
i.e.
3I 1 + 16I 2 = 20
Loop 2:
20 = 3I 1 + (I 1 - I 2 )(4 + 2)
i.e.
9I 1 - 6I 2 = 20
(2)
3 × equation (1) gives:
9I 1 + 48I 2 = 60
(3)
(1)
54I 2 = 40
Equation (3) – (2) gives:
from which,
I2 =
40
= 0.7407 A
54
Substituting in (1) gives:
3I 1 + 16(0.7407) = 20
i.e.
3I 1 = 20 – 11.851 = 8.149
from which,
I1=
8.149
= 2.716 A
3
i.e. the current flowing in the 3  resistor = 2.716 A
P.d. across the 10 Ω resistor = I 2 ×10 = 0.7407 ×10 = 7.407 V
P.d. across the 2 Ω resistor = (I 1 - I 2 )×2 = (2.716 - 0.7407) ×2
= 1.9753 ×2 = 3.951 V
5. For the network shown in circuit (b) below, find: (a) the current in the battery, (b) the current in
the 300  resistor, (c) the current in the 90  resistor, and (d) the power dissipated in the 150 
resistor.
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The currents are labelled as shown in the diagram below.
(a)
Loop 1:
8 = 20I 1 + (60 + 90)I 2
i.e.
20I 1 + 150I 2 = 8
Loop 2:
8 = 20I 1 + (I 1 - I 2 )(300 + 150)
i.e.
470I 1 - 450I 2 = 8
(2)
3 × equation (1) gives:
60I 1 + 450I 2 = 24
(3)
Equation (2) + (3) gives:
from which,
(1)
530I 1 = 32
I1=
32
= 0.06038 A
530
i.e. the current in the battery = 60.38 mA
(b)
Substituting in (1) gives:
20(0.06038) + 150I 2 = 8
i.e.
150I 2 = 8 – 1.2076 = 6.7924
404
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from which,
I2=
6.7924
= 0.04528 A = 45.28 mA
150
Hence, current in 300  resistor = I 1 - I 2 = 60.38 – 45.28 = 15.10 mA
(c)
The current in the 90  resistor = I 2 = 45.28 mA
(d)
The power dissipated in the 150  resistor = (I 1 - I 2 ) 2 × 150
= 15.10  103   150
2
= 0.0342 W or 34.20 mW
6. For the bridge network shown in circuit (c) below, find the currents I 1 to I 5
From loop 1:
From loop 2:
i.e.
6.6  3I A  2  I A  I B 
0 = 4  2  I A   5I B  3I A
5I A  2I B  6.6
(1)
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7I A  5I B  8
(2)
5  (1) gives:
25I A  10I B  33
(3)
2  (2) gives:
14I A  10I B  16
(4)
and
(3) – (4) gives:
= 49
39I A
IA 
and
Substituting in (1) gives:
49
= 1.256 A
39
5(1.256) + 2I B  6.6
IB 
from which,
Hence, correct to 2 decimal places,
6.6  5(1.256)
= 0.160 A
2
I1 = I A = 1.26 A
I 2 = 2 - I A = 2 – 1.256 = 0.74 A
I 3 = I B = 0.16 A
I 4 = I A + I B = 1.256 + 0.160 = 1.42 A
I 5 = 2 - I A - I B = 2 – 1.26 – 0.16 = 0.58 A
EXERCISE 175, Page 393
Answers found from within the text of the chapter, pages 388 to 393.
EXERCISE 176, Page 393
1. (a) 2. (d) 3. (c) 4. (b) 5. (c)
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