Question:
Published on: 21 June, 2022

Determine the value R in Fig. 15(a) such that 4 Ω resistor consumes maximum power.

Fig. 15(a)

Redraw the circuit of Fig. 15(a)

Fig. 15(b) Redraw the Fig. 15(a) to obtain equivalent resistance.

$$R_1=\frac{3\times R}{3+R+3}=\frac{3R}{6+R}$$

$$R_2=\frac{3\times3}{3+R+3}=\frac{9}{6+R}$$

$$R_3=\frac{3\times R}{3+R+3}=\frac{3R}{6+R}$$

The total internal resistance or equivalent resistance of the circuit is

$$R_i=\frac{(R_2+3)\times R_1}{(R_2+3+R_1)}+R_3$$

According to the maximum power transform theorem internal resistance equal to the load resistance for maximum power transferred to the load.

$$\frac{(R_2+3)\times R_1}{(R_2+3+R_1)}+R_3=4$$

Or $$\left(R_2+3\right)\times R_1+\left(R_2+3+R_1\right)R_3=4\left(R_2+3+R_1\right)$$

Or $$\left(\frac{9}{6+R}+3\right)\times\frac{3R}{6+R}+\left(\frac{9}{6+R}+3+\frac{3R}{6+R}\right)\frac{3R}{6+R}=4\left(\frac{9}{6+R}+3+\frac{3R}{6+R}\right)$$

Or $$\left(9+3(6+R)\right)\times\frac{3R}{6+R}+\left(9+3(6+R)+3R\right)\frac{3R}{6+R}=4\left(9+3(6+R)+3R\right)$$

Or $$\left(27+3R\right)\times\frac{3R}{6+R}+\left(27+6R\right)\times\frac{3R}{6+R}=4(27+6R)$$

Or $$\frac{3R}{6+R}\left(54+9R\right)=108+24R$$

Or $$\left(36+8R\right)\left(6+R\right)=9R(6+R)$$

Or $$\left(6+R\right)\left(36-R\right)=0$$

Or $$36-R=0$$

Or $$R=36\ \mathrm{\Omega}$$

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