In the circuit given below the current is to be measured. If the ammeter shown is a galvanometer with a resistance \( R_G = 60\,\Omega \) but converted to an ammeter by a shunt resistance \( r_s = 0.02\,\Omega \), and the cell has emf 3 V, then the flow of current is nearly :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.99 A
An ammeter made from a galvanometer has the galvanometer and the shunt in parallel, so its resistance is
\( \displaystyle R_A = \frac{R_G\,r_s}{R_G + r_s} = \frac{60\times0.02}{60 + 0.02} \approx 0.02\,\Omega \)
The ammeter is in series with the \( 3\,\Omega \) resistor, so the total resistance of the circuit (cell taken as ideal) is
\( \displaystyle R = 3 + 0.02 = 3.02\,\Omega \)
\( \displaystyle I = \frac{E}{R} = \frac{3\,\mathrm{V}}{3.02\,\Omega} = 0.99\,\mathrm{A} \)