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In the circuit shown in the figure the electromotive force of the battery is 9 V and its internal resistance is 15\Omega The two identical voltmeters can be considered ideal. Let V_{1} and V_{1}{}' reading of 1st voltmeter when switch is open and closed respectively. Similarly, V_{2} andV_{2}{}' be the reading of 2nd voltmeter when switch is open and closed respectively.\frac{V'_2-V_2}{V_1-V'_1}=

 

Option: 1

1


Option: 2

3


Option: 3

5


Option: 4

6


Answers (1)

best_answer

When switch is open : V_1=V_2=\frac{1}{2}\left(\frac{9}{15+10+20}\right) \times 30=3 V

When switch is closed : V_1^{\prime}=\frac{9}{15+10+20} \times 10=2 \mathrm{~V}

V_2^{\prime}=\frac{9}{15+10+20} \times 20=4 V            \therefore \frac{V_2^{\prime}-V_2}{V_1-V_1^{\prime}}=\frac{4-3}{3-2}=1

 

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Devendra Khairwa

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