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Thevenin's theorem
Thevenin's theorem states:
'The current in any branch of a network is that which would result if an e.m.f equal to the p.d.
across a break made in the branch, were introduced into the branch, all other e.mf 's being
removed and represented by the internal resistances of the sources.'
The procedure adopted when using Thevenin's theorem is summarised below.
To determine the current in any branch of an active network (i.e. one
containing a source of e.m.f.):
(i)
(ii)
(iii)
(iv)
I
E
R
r
remove the resistance R from that branch,
determine the open-circuit voltage, E, across the break,
remove each source of e.m.f. and replace them by their in resistances
and then determine the resistance, r, 'looking-in' at the break,
determine the value of the current from the equivalent circuit shown in
Figure 33, i.e. I = E
R +r
Figure 33
Problem 7. Use Thevenin's theorem to find the current flowing in the 10  resistor for the
circuit shown in Figure 34(a).
5
Following the above procedure:
(i)
(ii)
The 10  resistance is removed from the circuit as shown in
Figure 34(b)
There is no current flowing in the 5  resistor and current I1 is
given by:
I1 = 10
R1+R2
p.d.
10V
8
10
2
Figure 34a
= 10 = 1A
2+8
5
A
across R2 = I1R2 = 1 x 8 = 8 V
10V
Hence p.d. across AB, i.e. the open-circuit voltage across the break, E
= 8 V.
(iii) Removing the source of e.m.f. gives the circuit of Figure 34(c).
8
2
Resistance, r = R3 + R1R2 = 5 + 2 x 8
R1+R2
2+8
I1
B
Figure 34b
5
=5 +1.6 = 6.6 
A
(iv) The equivalent Thevenin's circuit is shown in Figure 34(d).
Current I = E = 8
= 8 = 0.482 A
R+r - 10+6.6
16.6
Hence the current flowing in the 10  resistor of Figure 34(a) is
0.482 A
8
r
2
Figure 34c
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B
I
8V
10 
6.6 
Figure 34d
Problem 8 For the network shown in figure determine the current in the 0.8  resistor.
5 
(i)
The 0.8  resistance is removed from the circuit
as shown in Figure 35a
(ii)
Current I1 =
12V
0.8 
4 
1
p.d.across 4 resistor = 4I1 = 4 x 1.2 = 4.8 V
Figure 35a
5 
Hence p.d. across AB, ie the open-circuit voltage across
AB E = 4.8V
A
12V
1
(iii)
Removing the source of em.f gives the circuit
shown in Figure 35a . The equivalent circuit of Figure
35c is shown in Figure 35d , from which,
4 
I1
Figure 35b
5 
resistance r = 4 x 6 = 24 = 2.4
4 + 6 10
B
A
(iii)
4 
r
The equivalent Thevenin circuit is shown in
Figure35 e from which,
Current I =
1
Figure 35c
12 = 12 = 1.2A
1 + 5 + 4 10
E = 4.8 = 1.5A
R + r 2.4 + 4.8
B
A
4.8V
5 +1
= 6
4 
r = 0.8
r
r = 2.4
Figure 35d
B
Figure 35e
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Problem 9.
Use theveninns theorem to determine the current I flowing in the 4  resistor
shown in Figure 36 . Find also the power dissipated in the 4  resistor.
E 1 = 4V
(i)
The 4  resistance is removed from the circuit
as shown in Figure 36
E2 = 2V
4 
1
2
(ii)
Current I1 =
E1 – E2
r1 + r2
= 4 - 2 = 2A
2 +1 3
p.d.across AB E = E1- I1r1 = 4 – 2/3 x 2 = 2 2/3 V
Figure 36
(iii)
Removing the source of em.f gives the circuit
shown in Figure 36 . The equivalent circuit of Figure
36b is shown in Figure 36c , from which,
A
E 1 = 4V
E2 = 2V
resistance r = 2 x 1 = 2 
2+1 3
I1
1 
2
(iii)
B
The equivalent Thevenin circuit is shown in
Figure 36 d from which,
Figure 36b
Current I =
A
2
r
1 
E =
R+r
2 2/3 = 0.571A
2/3 + 4
Power dissipated in the 4 resistor
P = I2R = 0.5712 x 4 = 1.304 W
B
Figure 36c
E =2 2/3 V
r = 4
r = 2/3
Figure 36 d
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Problem 11. A Wheatstone Bridge network is shown in Figure 37. Calculate the current
flowing in the 32  and its direction, using Thevenin's theorem. Assume the source of e.m.f.
to have negligible resistance.
C
R1 = 2
The 32  reistor is removed from the
circuit as shown in Figure
(ii)
The P.d. across A and C, VAC =
R2 = 14
R5 = 32
E = 54 V
(i)
A
B
R4 = 11
R3 = 3
R1 x E = 2 x 54 = 8.31 V
R1 + R4
2 + 11
D
The p.d. between B and C, VBC =
C
R1 =2 
R2 x E = 14 x 54 = 44.47 V
R2 + R3
14 + 3
R2 =14 
A
B
R4 =11 
R3 =3
Hence p.d. across A and B,
= 44.47- 8.31 = 36.16 V
D
2
C
Point c is at potential of 54 V. between C and A is
54 – 8.31 =45.69 V. Between C and B is a voltage
drop of 44.47 V. Hence the voltage at A is greater
than at B, current must flow in the direction A to B.
14
B
A
11
D
2
(iii)
replacing the source e.m.f. with a short
circuit gives the circuit shown in Figure (c). The
circuit is redrawn in and simplified as shown in
Figure (d), and (e) from which the resistance
between terminals A and B,
3
CD
A
r = 2 x 11 + 14 x 3 = 1.692 + 2.471 = 4.163
2 + 11 14+ 3
14

11 
D
3
(iv)
The equivalent Thevenin's circuit is
shown in Figure from which
I= E
= 36.16
= 1A
r + R5
4.163 + 32
r = 4.163
r = 32
Hence the current in the 32 resistor of Figure is
1A flowing from A to B.
E =36.16 V
Figure
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