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

(N/A) $Emf$ of the battery,$E = 10 \;V$
Internal resistance of the battery,$r = 3 \;\Omega$
Current in the circuit,$I = 0.5 \;A$
Let the resistance of the resistor be $R$.
Using Ohm's law for the entire circuit,$I = \frac{E}{R + r}$.
Rearranging for $R + r$,we get $R + r = \frac{E}{I} = \frac{10}{0.5} = 20 \;\Omega$.
Thus,$R = 20 - 3 = 17 \;\Omega$.
The terminal voltage $V$ of the battery is given by $V = E - Ir$ or $V = IR$.
Using $V = IR$,we get $V = 0.5 \;A \times 17 \;\Omega = 8.5 \;V$.
Therefore,the resistance of the resistor is $17 \;\Omega$ and the terminal voltage of the battery is $8.5 \;V$.