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  • As shown in the figure, the voltmeter reads 2 V across 5  resistor. The resistance of the voltmeter is _

As shown in the figure, the voltmeter reads 2 V across 5 \Omega resistor. The resistance of the voltmeter is ___  \Omega .

                                                 

Option: 1

20


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Method-I:

\begin{aligned} & R_{\text {eq }}=2+\frac{5 R}{5+R}=\frac{10+7 R}{5+R} \\ & i=\frac{3}{R_{e q}}=\frac{3(5+R)}{10+7 R} \\ & i_1=\frac{2}{5}, i_2=\frac{2}{R} \\ & i=i_1+i_2 \end{aligned}

\begin{aligned} & \frac{3(5+R)}{10+7 R}=\frac{2}{5}+\frac{2}{R}=\frac{2(5+R)}{5 R} \\ & 15 R(5+R)=2(5+R)(10+7 R) \\ & 75 R+15 R^2=2\left(50+35 R+10 R+2 R^2\right) \\ & 15 R^2+75 R=14 R^2+90 R+100 \\ & R^2-15 R-100=0 \\ & R=\frac{15 \sqrt{225 \times 1 \times 100}}{2} \end{aligned}

\begin{aligned} & =\frac{15 \pm \sqrt{625}}{2}=\frac{15 \pm 25}{2} \\ & \mathrm{R}=20 \Omega \end{aligned}

Method-II:

Given potential across 5 \Omega and voltmeter is 2V. To find resistance R of voltmeter.
Let current in 5 \Omega  be i1, and in R i2.

i_{1}=\frac{2}{5}   ,      i_{2}=\frac{2}{R}                    

i=i_{1}+i_2

\begin{aligned} & \frac{3(5+R)}{10+7 R}=\frac{2}{5}+\frac{2}{R}=\frac{2(5+R)}{5 R} \\ & 15 R(5+R)=2(5+R)(10+7 R) \\ & 75 R+15 R^2=2\left(50+35 R+10 R+2 R^2\right) \\ & 15 R^2+75 R=14 R^2+90 R+100 \end{aligned}

\begin{aligned} & R^2-15 R-100=0 \\ & R=\frac{15 \sqrt{225 \times 1 \times 100}}{2} \\ & =\frac{15 \pm \sqrt{625}}{2}=\frac{15 \pm 25}{2} \\ & R=20 \Omega \end{aligned}

 

Posted by

Kuldeep Maurya

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