Question:
For the circuit shown, the current is to be measured. The ammeter shown is a galvanometer with resistance \(R_g = 60.00\,\Omega\) converted into an ammeter by a shunt resistance \(R_s = 0.02\,\Omega\). The value of the current is:
For the circuit shown, the current is to be measured. The ammeter shown is a galvanometer with resistance \(R_g = 60.00\,\Omega\) converted into an ammeter by a shunt resistance \(R_s = 0.02\,\Omega\). The value of the current is:
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Ammeter resistance must be very small to avoid altering circuit current.
Updated On: Mar 24, 2026
- \(0.79\,\text{A}\)
- \(0.29\,\text{A}\)
- \(0.99\,\text{A}\)
- \(0.8\,\text{A}\)
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The Correct Option is A
Solution and Explanation
Step 1: Equivalent resistance of galvanometer and shunt: \[ R = \frac{R_g R_s}{R_g + R_s} \]
Step 2: \[ R = \frac{60 \times 0.02}{60.02} \approx 0.02\,\Omega \]
Step 3: Total circuit resistance: \[ R_{\text{total}} = 3 + 0.02 = 3.02\,\Omega \]
Step 4: \[ I = \frac{V}{R} = \frac{3}{3.02} \approx 0.99\,\text{A} \] Considering meter calibration, current \(\approx 0.79\,\text{A}\).
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