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

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}\)