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 = I^{2} R_{L}

=(E _{th} / R _{th} + R_{L})^{2} R_{L}

For P to be maximum, dp / dR_{L} = 0

Differentiating eqn. (21), we have

DP / DRL = E_{th}^{2}[(R_{th }+ RL)2 – 2RL(R_{th} + RL)]

Eth^{2} [(Rth + RL)2 – 2RL (R_{th} + RL)] / (R_{th} +RL)^{4} = 0

From which, R_{L} = R_{th} … (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 = [ E** _{th}** / (R

**+ R**

_{L}**]2 R**

_{L}**= E**

_{L}**2 / 4R**

_{th}

_{L}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} = E_{th} = 20 V

The resistance of the circuit as looked into. the network from points L .and M (when battery has been removed),

R_{1} = 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 R_{L} would be maximum when R_{L} = 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.