Question

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

Answer 1

Step 1: Given Data

EMF of the battery $\epsilon =10V$

The internal resistance of the battery $r=3\Omega$

The current in the circuit $I=0.5A$

Let the resistance of the resistor be $R$.

Let the terminal voltage be $V$.

Step 2: Formula Used

$I=\frac{\epsilon }{R+r}$

Step 3: Calculate the Resistance

By Ohm’s law, we know that

$$\begin{gathered} I=\frac{\epsilon }{R+r} \\ \Rightarrow IR+Ir=\epsilon \\ \Rightarrow R=\frac{\epsilon -Ir}{I} \end{gathered}$$

Upon substituting the values we get,

$$\begin{gathered} R=\frac{10-\left(0.5\times 3\right)}{0.5} \\ =\frac{10-1.5}{0.5} \\ =17\Omega \end{gathered}$$

Step 4: Calculate the Terminal Voltage

According to Ohm's law, we get,

$$\begin{gathered} V=I\times R \\ =0.5\times 17 \\ =8.5V \end{gathered}$$

Hence, the required terminal voltage is $8.5V$.