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)**

Answer:

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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