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. $V_2=-4V,I_3=\frac{5}{2}A$
• B. $V_2=+4V,I_3=\frac{5}{2}A$
• C. $V_2=-3V,I_3=1A$
• D. $V_2=+3V,I_3=2A$

Answer 1

Answer: (C)

Resistance $R_1$ and $R_4$ are connected in series, so the circuit can be redrawn as

As given, $I_1=0A$, so the potential drop of $2V$ takes place across the parallel branch $AB$ and resistor $R_3$.
Hence, $2=I_3{R}_3=I_{3}2$
$\Rightarrow I_3=1A$
Similarly, $V_{AB}=5-I_3{R}_2$
$2=5-1.R_2$
$\Rightarrow R_2=3\Omega$
$\Rightarrow V_2=-I_3{R}_2=-3\times 1=-3V$