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A galvanometer of resistance 60.00 Ω is used to measure current in a circuit with a 3 V source and a 3 Ω resistor. Used directly as the meter, the current reading is:
A0.99 A
B0.048 A
C1.00 A
D0.50 A
Answer & Solution
Correct answer: B. 0.048 A
1. The galvanometer adds its full resistance in series with the 3 Ω resistor.
2. Total resistance: $R_G + 3 = 60 + 3 = 63\ \Omega$.
3. Current: $I = \dfrac{V}{R} = \dfrac{3}{63}$.
4. $I = 0.048$ A — far below the true value, showing why a bare galvanometer disturbs the circuit.
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.161_
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