For the resistance network shown in the figure, choose the correct option(s)

For the resistance network shown in the figure, choose the correct option(s)

1. The current through PQ is zero.
2. $\mathrm{I}_{1}=3 \mathrm{~A}$
3. The potential at $S$ is less than that at $Q$.
4. $\mathrm{I}_{2}=2 \mathrm{~A}$

Solution

Resistance of arm $P Q$ and $S T$ becomes ineffective as $P \& Q$ and $S \& T$ are at the same potential. The equivalent circuit is as shown in the figure.
The resistance of the upper arm $R_{1}=2 \Omega+2 \Omega+2 \Omega=6 \Omega$
The resistance of the lower arm $R_{2}=4 \Omega+4 \Omega+4 \Omega=12 \Omega$ Equivalent resistance of the circuit, $\begin{array}{l} R_{\mathrm{eq}}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{(6 \Omega)(12 \Omega)}{6 \Omega+12 \Omega}=4 \Omega \\ \therefore I_{1}=\frac{12 \mathrm{~V}}{4 \Omega}=3 \mathrm{~A} \\ I_{2}=\left(\frac{12}{6+12}\right) \times 3=2 \mathrm{~A} \\ I_{3}=I_{1}-I_{2}=1 \mathrm{~A} \end{array}$ Potential difference across $A$ and $P$, $\begin{array}{l} V_{A}-V_{P}=I_{2} \times 2 \Omega=(2 A)(2 \Omega) \\ 12 V-V_{P}=4 V \text { or } V_{P}=8 V \end{array}$ Potential difference across $A$ and $Q$, $\begin{array}{l} V_{A}-V_{Q}=I_{3} \times 2 \Omega=(1 A)(4 \Omega) \\ 12 V-V_{Q}=4 V \\ V_{Q}=12 \mathrm{~V}-4 \mathrm{~V}=8 \mathrm{~V} \end{array}$ Potential difference across $P$ and $S$, $\begin{array}{l} V_{P}-V_{S}=(2 A)(2 W)=4 V \\ 8 V-V_{S}=4 V \Rightarrow V_{S}=4 V \\ \therefore V_{S} < V_{Q} \end{array}$ `

Asked in: JEE Advanced 2012 (Paper 1)

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