Question

If the resistances are chosen for the circuit shown in figure in such a way that no current flows through the battery with emf $E_{1}$, the voltage $V_{2}$ across $R_{2}$ and the current $I_{3}$ flowing through $R_{3}$ are respectively,

• A. $V_{2}=-4 V , I_{3}=\frac{5}{2} A$
• B. $V_{2}=+4 V , I_{3}=\frac{5}{2} A$
• C. $V_{2}=-3 V, I_{3}=1 A$
• D. $V_{2}=+3 V, I_{3}=2 A$

Answer 1

Answer: (C)

Resistance $R_{1}$ and $R_{4}$ are connected in series, so the circuit can be redrawn as

As given, $I_{1}=0 A$, so the potential drop of $2 V$ takes place across the parallel branch $AB$ and resistor $R_{3}$.
Hence, $2=I_{3} R_{3}=I_{3} 2$
$\Rightarrow \, I_{3}=1 A$
Similarly, $V_{A B}=5-I_{3} R_{2}$
$2=5-1 . R_{2}$
$\Rightarrow \, R_{2}=3 \Omega$
$\Rightarrow \, V_{2}=-I_{3} R_{2}=-3 \times 1=-3 V$