Question

A battery of emf $1.4V$ and internal resistance $2\Omega$ is connected to a resistor of $100\Omega$ resistance through an ammeter. The resistance of the ammeter is $\frac{4}{3}\Omega$. A voltmeter has also been connected to find the potential difference across the resistor. What is the resistance of the voltmeter if the ammeter reads 0.02A of current?

• A. $200\Omega$
• B. $400\Omega$
• C. $100\Omega$
• D. $50\Omega$

Answer 1

The correct option is C $100\Omega$

Let the resistance of the voltmeter is R

$\frac{1}{R_{BC}}=\frac{1}{100}+\frac{1}{R}{R}_{BC}=\frac{100R}{(R+100)}$

$R_{eq}=2+\frac{4}{3}+R_{BC}$

$\frac{V}{I}=\frac{10}{3}+\frac{100R}{(R+100)}$

$\frac{1.4}{0.02}=\frac{10}{3}+\frac{100R}{(R+100)}$

$\frac{100R}{(R+100)}=70-\frac{10}{3}$

$\frac{100R}{(R+100)}=\frac{200}{3}$

$\frac{R}{(R+100)}=\frac{2}{3}3R=2R+2003R-2R=200R=200\Omega$

Hence option 'C' is correct.