## Q. 1.P.50

In the following circuit the value of R required for the transfer of maximum power through load resistance of 10 Ω is

(a) 5 Ω (b) 15 Ω (c) 10 Ω (d) 20 Ω

## Verified Solution

Let the required resistance be $R_x$. Then

$R_{ eq }=\frac{10 \times R_x}{10+R_x}$

When $R=5 \Omega, R_{ eq }=3.33 \Omega$

$P_{5 \Omega}=I_5^2 \times 3.33, I_5=\frac{20}{R_{ eq }+5}$

I is maximum and power is maximum ≅ 50 W