Question

A battery of emf $E$ and internal resistance $r$ is connected across a resistance $R$. Resistance $R$ can be adjusted to any value greater than or equal to zero. A graph is plotted between the current passing through the resistance $\left(I\right)$ and potential difference across the termainals of the battery $\left(V\right)$. Maximum power developed across the resistance $R$ is

• A. $5\text{W}$
• B. $15\text{W}$
• C. $25\text{W}$
• D. $10\text{W}$

Answer 1

The correct option is A $5\text{W}$

We know that, terminal potential difference, $V=E-Ir$

From the graph, When $I=0$ then $V=10\text{V}$
We know that when current is zero, terminal potential difference is equal to the emf of the battery. So, $E=10\text{V}$ (or we can use the above equation also)

When $V=0$ then $I=2\text{A}$
Using the above equation, $V=E-Ir$
$0=10-2r$
$\therefore r=5\Omega$

For maximum power trasfer, external resistance should be equal to internal resistance of the battery.
So $R=r=5\Omega$



Current in the circuit, $i=\frac{10}{5+5}=1\text{A}$
Maximum power developed across $R$, $P=i^{2}R=1^2\times 5=5\text{W}$