Question

A cell of emf 8.0 V having a finite resistance of $4\Omega$ is connected to a load resistance of $x\Omega$ for maximum power transfer. The value of $x$ should be

• A. 0 ohm
• B. 4 ohm
• C. 3 ohm
• D. 2 ohm

Answer 1

The correct option is B 4 ohm

$P=i^2.x$
$={\left(\frac{\in }{4+x}\right)}^2.x$
for $P_{max},\frac{dP}{dx}=0$
$\frac{dp}{dx}=\frac{d}{dx}\left\{\frac{{\in }^2.x}{(4+x{)}^2}\right\}={\in }^2\frac{d}{dx}\left\{\frac{x}{(4+x{)}^2}\right\}$
$\frac{dp}{dx}=0=\frac{(4+x{)}^2.\frac{d}{dx}(x)-x\frac{d}{dx}(4+x{)}^2}{(4+x{)}^4}$
$0=(4+x{)}^2-2x(4+x)$
$0=(4+x)(4-x)$

for $x=4$, we get maximum power output.