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

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 \mathrm{S}_2 is also closed, then reading of ammeter becomes:

Option: 1

3 / 2 times the initial value 
 


Option: 2

 \frac{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:

Reading of ammeter = \frac{6}{R_A+R_V}=i_1

Reading of voltmeter =  \frac{6 R_V}{R_A+R_V}=V_1

On closing S_1: Reading of ammeter =\frac{6}{R_A+\frac{R R_v}{R+R_v}}=i_2=2 i_1

                      Reading of voltmeter =i_2\left(\frac{R R_v}{R+R_v}\right)=\frac{V_1}{2}

        \Rightarrow 2 i_1 \times \frac{R 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 \quad R_v=3 R

                 \mathrm{i}_2=2 \mathrm{i}_1 \Rightarrow \frac{6}{\mathrm{R}_{\mathrm{A}}+\frac{R R_{\mathrm{V}}}{\mathrm{R}+\mathrm{R}_{\mathrm{V}}}}=2\left(\frac{6}{\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{V}}}\right) \Rightarrow \mathrm{R}_{\mathrm{A}}=\frac{3 \mathrm{R}}{2}

\left(\because R_V=3 R\right)

On closing \mathrm{S}_2: \mathrm{i}_3=\frac{6}{\mathrm{R}_{\mathrm{A}}}=\frac{6}{\left(\frac{3 \mathrm{R}}{2}\right)}=\frac{4}{\mathrm{R}}

\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

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mansi

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