Question

The shunt resistance to be connected across a galvanometer of resistance $50,\Omega$ to send $\dfrac{1}{1000}^{\text{th}}$ of the main circuit current through the galvanometer is nearly

• A. $0.01,\Omega$
• B. $0.2,\Omega$
• C. $0.05,\Omega$
• D. $0.02,\Omega$
• E. $0.5,\Omega$

Hint:

Physics Tip: To convert galvanometer into ammeter, shunt resistance must be very small so that maximum current bypasses the meter.

Answer 1

The Correct Option is C

Concept:
A shunt resistance is connected in parallel with galvanometer so that most current bypasses the galvanometer. For galvanometer resistance $G$ and shunt $S$: $$I_g G = I_s S$$ where $I_g$ = galvanometer current, $I_s$ = shunt current.
Step 1: Given current division.
Only $\dfrac{1}{1000}$ of main current flows through galvanometer. Let total current = $I$ Then: $$I_g=\frac{I}{1000}$$ Remaining current through shunt: $$I_s=I-\frac{I}{1000}=\frac{999I}{1000}$$
Step 2: Apply parallel branch relation.
$$I_g G = I_s S$$ $$\frac{I}{1000}\times 50=\frac{999I}{1000}S$$ Cancel $\frac{I}{1000}$: $$50=999S$$ $$S=\frac{50}{999}\approx 0.05\Omega$$
Step 3: Final answer.
Hence required shunt resistance is $0.05\Omega$. Therefore option (C). :contentReference[oaicite:0]{index=0}