Question

Resistances $R_1$ and $R_2$ each $60\Omega$ are connected in series as shown in figure. The Potential difference between $A$ and $B$ is kept 120 volt. Then what will be the reading of voltmeter connected between the points $C\text{\&}D$ if resistance of voltmeter is $120\Omega$

• A. 48 V
• B. 24 V
• C. 40 V
• D. None

Answer 1

Answer: (A)

48 V

According to the above problem,
the voltmeter with resistance $R_v=120$ ohm and resistor $R_2=60$

ohm are in parallel, hence equivalent resistance between them is

$R_e=\frac{R_2{R}_v}{R_2+R_v}$

$R_e=\frac{\left(60\right)\left(120\right)}{60+120}=40\Omega$

Now the reading of voltmeter can be obtain by

$V_v=V\left(\frac{R_e}{R_1+R_e}\right)$

$V_v=120\left(\frac{40}{60+40}\right)$

$V_v=40\left(\frac{6}{5}\right)=48$ V