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In the circuit shown in figure

\mathrm{V}_{1} and \mathrm{V}_{2} are two voltmeters having resistances 6000 \Omega and 4000 \Omega respectively E.M.F. of the battery is 250 volts, having negligible internal resistance.
Two resistances \mathrm{R}_{1} and \mathrm{R}_{2} are 4000 \Omega and 6000 \Omega respectively. The reading of the voltmeters \mathrm{V_{1}} when Switch \mathrm{S} is open

Option: 1

125 \mathrm{V}


Option: 2

150 \mathrm{V}


Option: 3

100 \mathrm{V}


Option: 4

200 \mathrm{V}


Answers (1)

best_answer

When switch S is open
\mathrm{R_{1}} and \mathrm{R_{2}} are in series. Let their resistance be \mathrm{\mathrm{R}^{\prime}}
\mathrm{R^{1}=4000+6000=10000 \Omega}

The voltmeter are also in series.
Let their resistance be \mathrm{\mathrm{R}^{\prime \prime}}, then
\mathrm{\mathrm{R}^{\prime \prime}=6000+4000=10000 \Omega}

The resistance \mathrm{\mathrm{R}^{\prime}} and \mathrm{\mathrm{R}^{\prime \prime}} are connected in parallel. Their equivalent resistance is given by

\mathrm{\operatorname{Req}=\frac{R^{\prime} \times R^{\prime \prime}}{R^{\prime}+R^{\prime \prime}}=\frac{10000 \times 10000}{20000}}
\mathrm{=5000 \Omega}

\mathrm{\text{Current from battery }D=\frac{E}{\operatorname{Re} q}=\frac{250}{5000}}
Current \mathrm{i_{1}} in the voltmeter
\mathrm{branch =\frac{1}{2} \times \frac{1}{20}=\frac{1}{40} \mathrm{amp}}
\mathrm{\text{Potential difference across }\mathrm{V}_{1}=\frac{1}{20} \times 6000=150 \, volt}
\mathrm{\text{Potential difference across }\mathrm{V}_{2}=\frac{1}{40} \times 4000=100\, volt}
 

Posted by

Kuldeep Maurya

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