A galvanometer of resistance G shows full-scale deflection at current I_g. To convert it into an ammeter reading full-scale I (I > I_g), we connect
A{'text': 'A low resistance (shunt) in parallel', 'label': 'A'}
B{'text': 'A high resistance in parallel', 'label': 'B'}
C{'text': 'A high resistance in series', 'label': 'C'}
D{'text': 'A low resistance in series', 'label': 'D'}
Answer & Solution
Correct answer: A. {'text': 'A low resistance (shunt) in parallel', 'label': 'A'}
1. An ammeter must read a large current with small potential drop.
2. Adding a low-resistance shunt in parallel diverts the excess (I − I_g) through the shunt.
3. Only I_g still flows through the galvanometer coil, protecting it.
4. Shunt resistance S = G × I_g / (I − I_g).
_Source: NCERT Class 12 Physics, Ch 3 "Current Electricity", §3 measurement_
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