## Q. 1.P.52

Maximum power delivered to R in Ω by a 20 V DC source with 5 Ω internal resistance is

## Verified Solution

We have

$R=5 \Omega, I=\frac{20}{10}=2 A$

So, the power delivered to R is

$P_{ L }=I^2 R=4 \times 5=20 W$