In the figure given below, the current through the galvanometer with a resistance of \( R_{G} = 40\ \Omega \), is approximately,
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.046\ \mathrm{A} \)
Reading the figure: the 3 Ω resistor, the galvanometer and the 2 V cell form a single loop, so the galvanometer is in series with the resistor and carries the whole circuit current. The cell is taken as ideal (no internal resistance is given).
\( R_{\mathrm{total}} = 3\ \Omega + R_{G} = 3\ \Omega + 40\ \Omega = 43\ \Omega \)
\( \displaystyle I = \frac{V}{R_{\mathrm{total}}} = \frac{2\ \mathrm{V}}{43\ \Omega} \approx 0.046\ \mathrm{A} \)