Question

An ammeter and a voltmeter are connected in series to a battery with an emf $\epsilon =6.0V$. When a certain resistance is connected in parallel with the voltmeter, the reading of the voltmeter decreases $2.0times$, whereas the reading of the ammeter increase the same number of times. Find the voltmeter reading after the connection of the resistance.

• A. $2V$
• B. $4V$
• C. $8V$
• D. $18V$

Answer 1

The correct option is B $4V$

Case (1):- Initially, when R is not connected

the resistance of the ammeter is $=R$

Voltage drop across voltmeter $=V$

By applying Kirchhoff's voltage law,

$V=6-iR$. . . . . . .(1)

$i=\frac{6-V}{R}$. . . . . .(2)

Case (2):- When resistance connected parallel to the voltmeter,

$V_f=\frac{V}{2}$, $i_f=2i$

$v_f=6-i_{f}R$

$\frac{V}{2}=6-2iR$

Put the value of $i$ from equation (2), we get

$\frac{V}{2}=6-2R\left(\frac{6-V}{R}\right)$

$\frac{V}{2}=6-12+2V$

$V=2(2V-6)$

$V=4V-12$

$3V=12$

$V=\frac{12}{3}$

$V=4V$

The correct option is B.