In the circuit shown in the figure the electromotive force of the battery is 9 V and its internal resistance is 15 Ω. The two identical voltmeters can be considered ideal. Let V1 and V1' reading of 1st voltmeter when switch is open and closed respectively. Similarly, V2 and V2' be the reading of 2nd voltmeter when switch is open and closed respectively. \frac{V_{2}^{'} - V_{2}^{}}{V_{1}

Engineering

Physics

Optical Instrument

Question

In the circuit shown in the figure the electromotive force of the battery is 9 V and its internal resistance is 15 Ω. The two identical voltmeters can be considered ideal. Let V₁ and V₁' reading of 1^st voltmeter when switch is open and closed respectively. Similarly, V₂ and V₂' be the reading of 2^nd voltmeter when switch is open and closed respectively. $\frac{V_{2}^{'} - V_{2}^{}}{V_{1} - V_{1}^{'}} =$ ?

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Current in circuit

$I = \frac{1}{5} A = 0.2 A$

When switch is open

V₁ + V₂ = 6V ...... (1)
V₁ = V₂ = 3V
when switch closed
V₁' = 2V; V₂' = 4V

$\frac{V_{2}^{'} - V_{2}}{V_{1} - V_{1}^{'}} = \frac{4 - 3}{3 - 2} = 1$

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