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  • An ammeter and a voltmeter are initially connected in series to a battery of zero internal resistance. When switch  is closed the readin

An ammeter and a voltmeter are initially connected in series to a battery of zero internal resistance. When switch S_{1} is closed the
reading of the voltmeter becomes half of the initial, whereas the reading of the ammeter becomes double. If now switch S_{2} is
also closed, then reading of ammeter becomes

 

Option: 1

3/2 times the initial value
 


Option: 2

3/2 times the value after closing S_{1}
 


Option: 3

3/4 times the value after closing S_{1}
 


Option: 4

3/4 times the initial value


Answers (1)

best_answer

Before Closing:
\begin{aligned} & \text { Reading of ammeter }=\frac{6}{R_A+R_V}=i_1 \\ & \text { Reading of voltmeter }=\frac{6 R_V}{R_A+R_V}=V_1 \end{aligned}

\begin{gathered} \text { On closing } S_1 \text { : Reading of ammeter }=\frac{6}{R_A+\frac{R R_V}{R+R_V}}=i_2=2 i_1 \\ \text { Reading of voltmeter }=i_2\left(\frac{R R_V}{R+R_V}\right)=\frac{V_1}{2} \end{gathered}

\begin{gathered} \Rightarrow 2 i_1 \times \frac{R_V}{R+R_V}=\frac{V_1}{2} \quad \Rightarrow 2\left(\frac{6}{R_A+R_V}\right)\left(\frac{R R_V}{R+R_V}\right)=\left(\frac{6 R_V}{R_A+R_V}\right) \frac{1}{2} \Rightarrow R_V=3 R \\ i_2=2 i_1 \Rightarrow \quad \frac{6}{R_A+\frac{R R_V}{R+R_V}}=2\left(\frac{6}{R_A+R_V}\right) \Rightarrow R_A=\frac{3 R}{2} \quad\left(\because R_V=3 R\right) \end{gathered}

\: { On \: closing } \: S_2: i_3=\frac{6}{R_A}=\frac{6}{\left(\frac{3 R}{2}\right)}=\frac{4}{R}

\begin{aligned} & \mathrm{i}_2=2 \mathrm{i}_1=2\left(\frac{6}{\frac{3 \mathrm{R}}{2}+3 \mathrm{R}}\right)=\frac{8}{3 \mathrm{R}} \\ & \therefore \quad \mathrm{i}_3=\frac{3}{2} \mathrm{i}_2 \end{aligned}

 

Posted by

Rakesh

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