A battery of emf \( 3\,\mathrm{V} \) and internal resistance \( r \) is connected in series with a resistor of \( 55\,\Omega \). An ammeter of resistance \( 2\,\Omega \) connected in circuit reads \( 50\,\mathrm{mA} \). The value of \( r \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3\,\Omega \)
The battery, the resistor and the ammeter are all in series, so the emf drives the current through the total resistance of the circuit, including the internal resistance and the resistance of the ammeter (Ohm's law).
\( \displaystyle I = \frac{\varepsilon}{R + R_A + r}, \qquad I = 50\,\mathrm{mA} = 0.05\,\mathrm{A} \)
\( \displaystyle R + R_A + r = \frac{3}{0.05} = 60\,\Omega \)
\( \displaystyle r = 60 - 55 - 2 = 3\,\Omega \)