A battery of emf 3 V and internal resistance \( r \) is connected in series with resistance of \( 55\,\Omega \) through an ammeter of resistance \( 1\,\Omega \). The ammeter reads the current 50 mA. The internal resistance \( r \) of the battery is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4\,\Omega \)
The battery, the \( 55\,\Omega \) resistor and the ammeter are in series, so by Ohm's law for the whole circuit the emf drives the current through the total resistance, including the internal resistance.
\( \displaystyle I = \frac{E}{R + R_A + r} \)
\( I = 50\,\mathrm{mA} = 0.05\,\mathrm{A} \)
\( E = 3\,\mathrm{V} \)
\( \displaystyle R + R_A + r = \frac{3}{0.05} = 60\,\Omega \)
\( \displaystyle r = 60 - 55 - 1 = 4\,\Omega \)