An electrical circuit contains a battery of 4 V, and a resistance of \( 2\,\Omega \). A galvanometer of resistance \( 40\,\Omega \) shunted by \( 10\,\Omega \) is used to measure current in the circuit. The value of current in the circuit will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.4 A
The galvanometer and its shunt are in parallel, and this combination is in series with the \( 2\,\Omega \) resistance, so the meter resistance must be included in the circuit.
\( \displaystyle R_{\mathrm{meter}} = \frac{G S}{G + S} = \frac{40 \times 10}{40 + 10} = 8\,\Omega \)
\( R_{\mathrm{total}} = 2\,\Omega + 8\,\Omega = 10\,\Omega \)
\( \displaystyle I = \frac{V}{R_{\mathrm{total}}} = \frac{4\,\mathrm{V}}{10\,\Omega} = 0.4\,\mathrm{A} \)