StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A milliammeter has a range of 10 mA and internal resistance of 9 ohm. It is connected in a circuit where additional resistors are present (as shown in a figure). When terminals A and B are used (current enters at A, leaves at B, terminal C is isolated), the meter gives full-scale deflection at current I. Given standard circuit: the milliammeter (9 ohm, 10 mA full scale) has a 1 ohm shunt in parallel, and a 90 ohm resistor in series with A, with C being a tap between the shunt and series resistor. Find the current I for full-scale deflection through A-B.

  1. (A) 100 mA
  2. (B) 900 mA
  3. (C) 1 A
  4. (D) 1.1 A

Correct answer: (D) 1.1 A

Solution

This is a standard multi-range ammeter/voltmeter problem. The circuit (standard for this JEE problem) has: galvanometer G = 9 ohm, full scale 10 mA, connected with a shunt of 1 ohm in parallel, and a series resistance of 90 ohm between terminal A and the parallel combination, with C as the junction between the 90 ohm resistor and the parallel combination, and B at the other end of the shunt. When A-B is used (C isolated): total resistance = 90 + (9*1)/(9+1) = 90 + 0.9 = 90.9 ohm. For 10 mA through G: voltage across parallel combination = 10 mA * 9 = 0.09 V. Current through shunt = 0.09/1 = 90 mA. Total current I = 10 + 90 = 100 mA. That gives option (A). Alternatively, if shunt = 0.1 ohm: Vg = 10e-3 * 9 = 0.09 V, I_shunt = 0.09/0.1 = 900 mA, I_total = 910 mA approx 900 mA or could be 1 A or 1.1 A depending on series R. The answer 100 mA with a 1-ohm shunt (no series R for A-B path) is the most consistent: I = 100 mA = option (A).

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →