This theorem is particularly useful for analysing communication networks. It is stated as follows:
“Maximum power output is obtained from a network when the load resistance is equal to the output resistance of the network as seen from the terminals of the load”.
Any network can be converted into a single voltage source by the use of Thevenin’s theorem (Fig. 99). The maximum power transfer theorem aims at R th 1 finding RL such that the power dissipated in RL is maximum.
P = I2 RL
=(E th / R th + RL)2 RL
For P to be maximum, dp / dRL = 0
Differentiating eqn. (21), we have
DP / DRL = Eth2[(Rth + RL)2 – 2RL(Rth + RL)]
Eth2 [(Rth + RL)2 – 2RL (Rth + RL)] / (Rth +RL)4 = 0
From which, RL = Rth … (221
It is worth noting that under these conditions the voltage across the load is half the open-circuit voltage at the terminals L and M.
Maximum power, P max = [ Eth / (RL + RL]2 RL = E th2 / 4RL
The process of adjusting the load resistance for maximum power transfer is called ‘load matching’. This is done in the following typical cases:
(i) Motor cars- here starter motor is matched to the battery.
(ii) Telephone lines and TV aerial leads-these are matched to the telephone instrument and TV receiver respectively.
Example 23. For the circuit shown in Fig. 100, find the current through RL when it takes on values of 5 n and 25 Q. Also, calculate the value of RL for which the power dissipated in it would be maximum and find this power.
Solution. The open circuit voltage V oc (also called Thevenin’s voltage Eth) which appears across terminals L and M is equal to the voltage drop across 10 n resistance.
Current flowing through the circuit EPQ = 60 / 20 x 10 = 2 A
Voltage drop over 10 Q resistance = 2 x 10 = 20 V
Hence V oc = Eth = 20 V
The resistance of the circuit as looked into. the network from points L .and M (when battery has been removed),
R1 = R th = 30 + 10 || 20 = 30 + 10 x 20 / 10 + 20 = 36.67Q
The whole circuit up to LM can now be replaced by a single source of e.m.f. and single resistance as shown· in Fig. 101.
(i) When RL = 5 Q, I = E th / R th + RL = 20 / 36.67 + 5 = 0.48 A
(ii) When RL = 25 Q, I = 20 / 36.67 + 25 = 0.324 A
According to the maximum power transfer theorem, power drawn by RL would be maximum when RL = R; or when RL = 36.67 Q
. . Maximum power drawn by RL = I2RL = ( Eth / R th + RL )2 RL
= (E th / RL + RL)2 RL = Eth / 4 RL
= 202 / 4 x 36.67 = 2.72 W.