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 \(R=36\ \mathrm{\Omega}\)
For the circuit shown below, find the potential difference between a and d:
Derive the expression of quality factor of series R-L-C circuit at resonance.
State Channel capacity theorem
What is meant by entropy of a source?