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A galvanometer of resistance 25 Ω is connected to a battery of 2 V along with a resistance of 3000 Ω . In this case, a full-scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 10 units , how much more resistance (in Ω ) should be added to the circuit in series?

Correct answer

6050

Step-by-step solution

The deflection in a galvanometer is directly proportional to the current. First case: 30 x = 2 25 + 3000 Second case: 10 x = 2 25 + 3000 + R ⇒ 3 = 25 + 3000 + R 25 + 3000 On solving, we get R = 6050 Ω

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