For the resistance network shown in the figure, choose the correct option(s). [2012]

Question

Smart Achievers · Accepted Answer

(1, 2, 3, 4)

Resistance of arm PQ and ST 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

$R_{1} = 2 Ω + 2 Ω + 2 Ω = 6 Ω$

The resistance of the lower arm

$R_{2} = 4 Ω + 4 Ω + 4 Ω = 12 Ω$

Equivalent resistance of the circuit

$R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{(6 Ω) (12 Ω)}{6 Ω + 12 Ω} = 4 Ω$

$∴$ $I_{1} = \frac{12 V}{4 Ω} = 3 A$

$I_{2} = (\frac{12}{6 + 12}) \times 3 = 2 A$

$I_{3} = I_{1} - I_{2} = 3 - 2 = 1 A$

Potential difference across A and P

$V_{A} - V_{P} = I_{2} \times 2 Ω = (2 A) (2 Ω)$

$∴$ $12 V - V_{P} = 4 V$ or $V_{P} = 8 V$

Potential difference across A and Q

$V_{A} - V_{Q} = I_{3} \times 4 Ω = (1 A) (4 Ω)$

$12 V - V_{Q} = 4 V$

$V_{Q} = 12 V - 4 V = 8 V$

Potential difference across P and S

$V_{P} - V_{S} = (2 A) (2 Ω) = 4 V$

$8 V - V_{S} = 4 V$ $\Rightarrow V_{S} = 4 V$

$∴$ $V_{S} < V_{Q}$

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