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Maximum Power Transfer Theorem
The maximum power transfer theorem states: 'The power transferred from a supply source to
a load is at its maximum when the resistance of the load is equal to the internal resistance of
the source.’
Hence, in Figure 1 when R = r the power
transferred from the source to the load is a
maximum.
6V
RL
R= 2.5
Figure 1
Problem 20. The circuit diagram of Figure 1 shows dry cells of source e.m.f. 6 V, and
internal resistance 2.5. If the load resistance RL is varied from 0 to 5  in 0.5  steps,
calculate the power dissipated by the load in each case. Plot a graph of R L (horizontally)
against power (vertically) and determine the maximum power dissipated.
When RL = 0, current I = E = 6 = 2.4 A and power dissipated in RL, P = I2RL,
r + RL 2.5
i.e. P = (2.4)2(0) = 0 W
When RL = 0.5  current I = E =
6
= 2A and power dissipated in RL, P = I2RL,
r + RL 2.5+0.5
i.e. P = (2)2(0.5) = 2 W
When RL = 1.0  current I = E =
6
= 1.714A and power dissipated in RL, P = I2RL,
r + RL 2.5+1.0
i.e. P = (1.714)2(1.0) = 2.94 W
With similar calculations the following table is produced:
Power (w)
RL()
I= E
r+ RL
P=I2RL (W)
0
2.4
0.5
2.0
1.0
1.5
1.5
2.0
1.33
2.5
1.2
3.0
1.09
3.5
11.0
4.0
4.5
1.714
0.923
0.857
5.0
0.8
0
2.00
2.94
3.38
3.56
3.60
3.57
3.50
3.41
3.31
3.20
4
3.5
3
2.5
2
1.5
1
0.5
0
A graph of RL against P is shown in Figure 2. The
maximum value of power is 3.60 W which
occurs when RL is 2.5 , i.e. maximum power
occurs when RL = r, which is what the maximum
power transfer theorem states.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
RL (ohms)
Figure 2
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By Calculus
Problem 21. A d.c. source has an open-circuit voltage of 30 V and an internal resistance of
1.5 . State the value of load resistance that gives maximum power dissipation and determine
the value of this power.
30 V
The circuit diagram is shown in Figure 3. From the
maximum power transfer theorem, for maximum
power dissipation, RL= r
I
RL
R= 1.5
Figure 3
When RL = 1.5  current I = E = 30 = 10 A and power dissipated in RL, P = I2RL,
r + RL 1.5+1.5
i.e. P = (10)2(1.5) = 150 W
Problem 22. Find the value of the load resistor RL shown in Figure 4(a) that gives maximum
power dissipation and determine the value of this power.
Using the procedure for Thevenin's theorem:
15V
RL
12
A
15V
12
3
RL
12V
12
3
(b)
(c)
(i)
Resistance RL is removed from the circuit as shown in Figure 4(b).
(ii)
The p.d. across AB is the same as the p.d. across the 12  resistor.
Hence E =
RL
2.4
B
B
Figure 4(a)
3
A
(d)
12 (15) = 12V
12 + 3
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(iii)
Removing the source of e.m.f. gives the circuit of Figure 4(c),
from which resistance, r = 12 x 3 = 36 = 2.4
12 + 3 15
(iv)
The equivalent Thevenin’s circuit supplying terminals AB is shown in Figure 4(d),
from which, current, I = E/(r + RL)
For maximum power, RL = r = 2.4 . Thus current, I =
12
= 2.5A.
2.4 + 2.4
Power, P, dissipated in load RL, P =I2RL - (2.5)2(24) = 15W
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