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A galvanometer of resistance 50 Ω is connected to battery of 3V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
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- 5050 Ω
- 5550 Ω
- 6050 Ω
- 4450 Ω
Correct Option: D
Total internal resistance = (50+2950) Ω
= 3000 Ω
Emf of the cell, ε = 3V
| ∴ Current = | = | = 1 × 10-3A = 1.0 mA | ||
| R | 3000 |
∴ Current for full scale deflection of 30 divisions is 1.0 mA.
∴ Current for a deflection of 20 divisions,
| I = |  | × 1 |  | mA or I = | mA | ||
| 30 | 3 |
Let the resistance be x Ω. Then
| x = | = | = | Ω | ||||||||
| I | | × 103A | | 2 | |||||||
| 3 | |||||||||||
= 4500 Ω
But the resistance of the galvanometer is 50Ω.
∴ Resistance to be added = (4500 –50) Ω = 4450 Ω
