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  •  In the circuit shown in figure <img alt="\mathrm{V_{1}}" src=

 In the circuit shown in figure

\mathrm{V_{1}} and \mathrm{V_{2}} are two voltmeters having resistances \mathrm{6000 \Omega} and \mathrm{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 \mathrm{4000 \Omega} and \mathrm{6000 \Omega} respectively. The reading of the voltmeters \mathrm{V_{1}} when Switch S is Closed  

Option: 1

125 \mathrm{V}


Option: 2

150 \mathrm{V}


Option: 3

100 \mathrm{V}


Option: 4

125 \mathrm{V}


Answers (1)

best_answer

When switch \mathrm{S} is closed
The circuit redrawn in this case is shown in figure. In this case \mathrm{\mathrm{V}_{1}} and \mathrm{\mathrm{R}_{1}} are in parallel.


Equivalent resistance of \mathrm{V}_{1} and \mathrm{R}_{1}
\mathrm{R^{\prime}=\frac{6000+4000}{6000+4000}=2400 \Omega}

Similarly for \mathrm{\mathrm{R}_{2}} and \mathrm{\mathrm{V}_{2}}
\mathrm{\mathrm{R}^{\prime \prime} \quad=\frac{6000 \times 4000}{6000+4000}=2400 \Omega}

So, the two equal resistances are connected in series
Hence reading of \mathrm{V_{1}=125 \mathrm{volt}}
And reading of \mathrm{\mathrm{V}_{2}=125 \mathrm{volt}}

 

Posted by

Kuldeep Maurya

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