Question

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

Answer 1

Step 1: Draw a diagram from the given information

Step 2 : Find the resistance of the resistor
Formula used: $I=\frac{E}{R+r}$
Given,
Emf of the battery, $E=10V$

Internal resistance of the battery, $r=3\Omega$

Current in the circuit, $I=0.5A$

Let the connected resistance be ${}^{\prime }R‘$

Then, the relation for current using Ohm's law will be,

$I=\frac{E}{R+r}\Rightarrow R+r=\frac{E}{I}\Rightarrow R=\frac{E}{I}-r$

Putting the values, we get,

$R=\frac{10}{0.5}-3=20-3=17\Omega$

$R=17\Omega$

Step 3 : Find the terminal voltage

Formula used: $V=IR$

Let the terminal voltage (voltage across the resistor), when the
circuit is closed, be $V$. Then, according to Ohm's law,

$V=IR$

$V=0.5\times 17=8.5V$

Final Answer:

The resistance is $17\Omega$ and the terminal voltage is$8.5V.$