Question

An ammeter and a voltmeter are connected in series to a battery with an $emfE=6$ 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 will be voltmeter reading before the connecting of the resistance?

• A. $1V$
• B. $2V$
• C. $3V$
• D. $4V$

Answer 1

Answer: (D)

Suppose $R_A=$ Resistance of ammeter,
$R_v=$ resistance of voltmeter
In the first case current is the circuit
$i=\frac{6}{\left(R_A+R_V\right)}\dots$(i)
And voltage across voltmeter $V=6$ - Voltage across
ammeter
$V=6-i{R}_A;V=6-\frac{6}{\left(R_A+R_V\right)}{R}_A\dots$(ii)
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 $V^{\prime }=6-i{R}_A$
$V^{\prime }=6-\frac{12}{\left(R_A+R_V\right)}$
It is given that
$V^{\prime }=\frac{V}{2}$
$6-\frac{12R_A}{\left(R_A+R_V\right)}$
$=\frac{\left[6-\frac{6}{R_A+R_V}\right]}{2}$
[From (ii) and (iii)]
$\frac{R_A}{R_A+R_V}=\frac{1}{3}$
$\Rightarrow \frac{R_V}{R_A}=2$
Substituting this value into equation (ii).
$V=4$