Question

Three resistors of resistances $100\Omega$, $200\Omega$ and $400\Omega$ are connected in series with a cell of emf $7.8V$ and negligible internal resistance. A voltmeter of resistance $200\Omega$ connected across the two resistors $200\Omega$ and $400\Omega$ separately reads $V_1$ and $V_2$ respectively. Then $V_2-V_1$ is equal to:

• A. $4.4V$
• B. $3.3V$
• C. $2.2V$
• D. $1.1V$

Answer 1

The correct option is D $1.1V$

When the voltmeter is connected across 200Ω200Ω resistance, current in the circuit is $\frac{7.8}{600}=0.013A$ and current through $200\Omega$ resistance is $0.0065A$.

$V_1=0.0065\times 200=1.3V$

$V_2=.006\times 400=2.4V{V}_2-V_1=1.1V$