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**Maximum Power Transfer Theorem **

Whenever the load is connected across a voltage source, power is transferred or moved from the source to the load. The amount of power transferred will depend upon the resistance of the load. If the resistance of load RL is made equal to the internal resistance Ri of any source, then maximum power is transferred or moved to the load RL. This is called as **maximum power transfer theorem** and can be defined as follows: Maximum power is transfer from a source to a load when the load resistance and source resistance are equal. This theorem applies the same to direct current as well as alternating current power.* To prove this theorem numerically, consider a voltage source of producing voltage E and internal resistance Ri and transferring power to a load resistance RL [See Fig. 1.20 (i)]. The current I passes through the circuit is given by :

For this source, produced voltage E and internal resistance Ri are constant. Therefore, the power transferred to the load depends upon RL. In order to determine the value of RL for which the value of P is maximum, it is a must to differentiate eq. (i) w.r.t. RL and set the result equal to 0.

Since (RL+Ri) cannot be zero,

∴ Ri− RL= 0

or RL= Ri

i.e. Load resistance = Internal resistance

Thus, for maximum power delivered, load resistance RL and internal resistance Ri of the source must be equal. Under that situation, the load is said to be matched to the source. Fig. 1.20 (ii) shows a graph of power transferred to RL as a function of RL. It may be noted that the efficiency of maximum power delivery is *50% as one-half of the total produced power is consumed in the internal resistance Ri of the source.

**Maximum Power Transfer Theorem** **Applications.**

Electric power systems never work for maximum power delivered because of very low efficiency and a huge voltage drop between generated voltage and load. However, in the electronic circuits, the maximum power delivered is normally desirable. For instance, in a public address system, it is desirable to have a load (i.e. speaker) “matched” to the amplifier circuit so that there is the maximum deliverance of power from the amplifier circuit to the speaker. In such condition, **efficiency is **sacrificed at the cost of high power transfer.**

Example 1.5. A generator produced 200 V and has an internal resistance of 100 Ω. Find the power transferred to a load of (i) 100 Ω (ii) 300 Ω. Comment on the result.

Solution.

Generated voltage, E = 200 V

Internal resistance, Ri= 100 Ω

(i) When load RL= 100 Ω

Thus, out of 200 W power generated by the generator, only 100W has reached the load i.e.

efficiency is 50% only

Thus, out of 100 watts of power developed by the generator, 75 watts is delivered to the load i.e.

efficiency is 75%.

Comments. Although in the case of RL = Ri a huge power (100 W) is delivered to the load, there is a large wastage of power in the generator. On the other hand, when RL and Ri are not equal.

Electronic devices generate small power. Therefore, if too much efficiency is sought, a huge number of such devices will be connected in series to get the required output. This will distort the output as well as increase the cost and size of the device.

Power transfer is less (75 W) but a smaller part is wasted in the generator i.e. efficiency is very high. Thus, it depends upon a particular condition as to what the load should be. If we want to deliver maximum power (e.g. in amplifiers) irrespective of efficiency, we should make RL = Ri, However, if efficiency is more necessary (e.g. in power systems), then internal resistance of the source should be taken smaller than the load resistance.

**Example 1.6. An audio amplifier generates an alternating output of 12 V before the connection to a load. The amplifier circuit has an equivalent resistance of 15 Ω at the output. What resistance the load required to have to generate maximum power? Also, finf the power output under this condition.**

**Solution**. In order to generate maximum power, the load (e.g. a speaker) should have a resistance of 15 Ω to match the amplifier circuit. The equivalent circuit is shown in Fig. 1.21.

∴ Load required, RL = 15 Ω

**Example 1.7. For the alternating current the generator is shown in ****Fig. 1.22 (i)****, find the value of load so that maximum power is delivered to the load (ii) the value of maximum power.**

(i) In the alternating current system, maximum power is transferred to the load impedance (ZL) when the load impedance is conjugate of the internal impedance (Zi) of the source. Now in the problem, Zi = (100 + j50)Ω.

For maximum power delivered, the load impedance should be conjugate of internal impedance i.e. ZL should be (100 − j50) Ω. This is shown in the dotted line in Fig. 1.22 (ii).

Point out that by making internal impedance and load impedance conjugate, the reactive terms cancel. The circuit then consists of internal and external resistances only. This is quite logical because power is only dissipated in resistances as reactances (XL or XC) dissipated no power

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Reference : Principles Of Electronics Multicolor Illustrative Edition By V K Mehta And Rohit Mehta