Question

A battery of emf $10\text{V}$ and internal resistance $3\Omega$ is connected to a resistor. If the current in the circuit is $0.5\text{A}$, what is the resistance of the resistor and terminal voltage of the battery when the circuit is closed ?

• A. $14\Omega ,5.5\text{V}$
• B. $16\Omega ,6.5\text{V}$
• C. $16\Omega ,7.5\text{V}$
• D. $17\Omega ,8.5\text{V}$

Answer 1

The correct option is D $17\Omega ,8.5\text{V}$

Given:
$E=10\text{V};r=3\Omega ;I=0.5\text{A}$

Using the relation,
$I=\frac{E}{R+r}$, we get

$R=\frac{E}{I}-r=\frac{10}{0.5}-3$

$R=17\Omega$

From Ohm's law,

$V=IR$

$\Rightarrow V=0.5\times 17=8.5\text{V}$

Therefore, option $\left(d\right)$ is correct.

Why this question? Key concept : Battery acting as a source (or battery is discharging) $V_A-V_B=\epsilon -ir$ $\Rightarrow$it is also called terminal voltage.