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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 $Y_3$ flowing through $R_3$ are respectively,
Options:
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Verified Answer
The correct answer is:
$V_2=-3 V_1 f_3=1 \mathrm{~A}$
Resistance $R_1$ and $R_4$ are connected in series, so the circuit can be redrawn as
As given, $I_1=0 \mathrm{~A}$, so the potential drop of $2 \mathrm{~V}$ takes place across the parallel branch $\mathrm{AB}$ and resistor $R_3$.
Hence, $2=I_3 R_3=I_3 2$
$$
\begin{array}{lc}
\Rightarrow I_3=1 \mathrm{~A} \\
\text { Similarly, } V_{A B}= 5-I_3 R_2 \\
2=5-1 . R_2 \\
\Rightarrow R_2=3 \Omega \\
\Rightarrow V_2=-I_3 R_2=-3 \times 1=-3 \mathrm{~V}
\end{array}
$$
Hence, the correct option is (3).
As given, $I_1=0 \mathrm{~A}$, so the potential drop of $2 \mathrm{~V}$ takes place across the parallel branch $\mathrm{AB}$ and resistor $R_3$.
Hence, $2=I_3 R_3=I_3 2$
$$
\begin{array}{lc}
\Rightarrow I_3=1 \mathrm{~A} \\
\text { Similarly, } V_{A B}= 5-I_3 R_2 \\
2=5-1 . R_2 \\
\Rightarrow R_2=3 \Omega \\
\Rightarrow V_2=-I_3 R_2=-3 \times 1=-3 \mathrm{~V}
\end{array}
$$
Hence, the correct option is (3).
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