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  • An ammeter and a voltmeter are connected in series to a battery with an emf E = 6 volt when a certain resistance is connected in parallel with voltmeter, the reading of latter decreases t

An ammeter and a voltmeter are connected in series to a battery with an emf E = 6 volt when a certain resistance is connected in parallel with voltmeter, the reading of latter decreases two times, where as the reading of the ammeter increasing the same number of times.What is ratio of resistance of voltmeter to resistance of ammeter

Option: 1

2


Option: 2

1/2


Option: 3

1/3


Option: 4

3


Answers (1)

best_answer

Suppose R_A= Resistance of ammeter, R_V= resistance of Voltmeter In the first case current is the circuit
i=\frac{6}{\left(R_A+R_V\right)}\quad \quad \quad \quad \dots(1)
And voltage across voltmeter V=6 - Voltage across ammeter
\begin{aligned} & \mathrm{V}=6-\mathrm{iR}_{\mathrm{A}} \\ & V=6-\frac{6}{\left(R_A+R_V\right)} R_A \quad \quad \quad \quad \dots(2) \end{aligned}
In the second case reading of ammeter becomes two times i.e. the total resistance become half while the resistance of ammeter remains unchanged. Hence
i=\frac{6}{\left(R_A+R_V\right) / 2}=i=\frac{12}{\left(R_A+R_V\right)}
and voltage across voltmeter V^{\prime}=6-i R_A

V^{\prime}=6-\frac{12}{\left(R_A+R_V\right)} \quad \quad \quad \quad \dots(3)

It is given that
V^{\prime}=\frac{V}{2} \quad \quad \quad \quad \dots(4) \\
\begin{gathered} 6-\frac{12 R_A}{\left(R_A+R_V\right)}=\frac{\left[6-\frac{6}{R_A+R_V}\right]}{2} \text { [From (2) and (3)] } \\ \frac{R_A}{R_A+R_V}=\frac{1}{3} \Rightarrow \frac{R_V}{R_A}=2 \end{gathered}

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sudhir kumar

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