Question

In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is

• A. 1 nF
• B. 1 $\mu F$
• C. 1 mF
• D. 10 mF

Answer 1

The correct option is D 10 mF



The frequency at which the load is resistive and it is to $0.5\Omega$ i.e. The load is resistive means, the imaginary part of the load is equal to zero and real part is equal to 0.5$\Omega$

$Z_{load}=\frac{1\times \frac{1}{Cs}}{1+\frac{1}{Cs}}+Ls$

$=\frac{1}{1+Cs}+Ls$

$=\frac{(1-Cs)}{1-C^2{S}^2}+Ls$

$PutS=j\omega$;

$Z_{load}=\frac{1-j\omega C}{1+C^2{\omega }^2}+j\omega L$

$=\frac{1}{1+C^2{\omega }^2}+j\omega (L-\frac{C}{1+C^2{\omega }^2})$

Real part of the

$Z_{load}=\frac{1}{1+C^2{\omega }^2}$

$Z_{load}=\frac{1}{1+{\omega }^2{C}^2}=0.5$

Putting, $\omega =100rad/s$

we get, C =10 mF