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Solution

Effective resistance between the coltmeter and 1000 Ω resistor \frac{1}{R_{E F F}} = \frac{1}{1000} + \frac{1}{1000} = \frac{2}{1000} R_{E F F} = 500 Ω The equivalent resistance of the Voltmeter (1000 Ω) and the given resistance of 1000 Ω is 500 Ω . Total effective resistance in the circuit R = 500 + 500 = 1000 Ω Current in circuit, I = \frac{V}{R} = \frac{1.5}{1000} = 1.5 \times 10^{- 3} A Potential drop across 1000 Ω resistor and voltmeter V_{500} = I R_{E F F} = 1.5 \times 10^{- 3} \times 500 = 0.75 V