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An ideal battery of EMF 1.3 V is connected to a Wheatstone-bridge-type circuit assembled with three identical high-resistance voltmeters (resistance R_V each) and two identical milliammeters (resistance R_A each, with R_V >> R_A). One milliammeter reads three times the other. Find which statements about voltmeter readings are correct.

1. Reading of Voltmeter A is 1.0 V
2. Reading of Voltmeter B is approximately 0.62 V
3. Reading of Voltmeter C is 0.4 V
4. Reading of Voltmeter C is approximately 0.68 V

Correct answer: Reading of Voltmeter A is 1.0 V

Solution

With three voltmeters (A, B, C) and two milliammeters in a bridge configuration, and R_V >> R_A, the voltmeters draw negligible current. If one milliammeter carries I and the other 3I, then by current division the resistances of their branches relate as 3:1. The total voltage 1.3 V divides between the two branches. Using KVL and the 1:3 current ratio: the branch with current I has resistance 3R_A (relative), and the branch with 3I has resistance R_A. Voltage across the high-current branch = 3I * R_A, voltage across low-current branch = I * 3R_A. For voltmeter readings: V_A reads the terminal voltage or across a specific branch. If V_A is across the full battery minus one ammeter: V_A ~ 1.0 V (given R_A << R_V, voltmeter A in parallel with one main branch reads approximately 1.0 V).

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