Resistance of $6 o h m$ each are connected in the manner shown in adjoining figure.With the current $0.5$ ampere as shown in figure, the potential difference $V_{P} - V_{Q}$ is :

3.6 V

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6.0 V

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3.0 V

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7.2 V

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Solution

The correct option is C $3.0 V$
From the figure, the net resistance between PQ is,
$R_{n e t} = \frac{6}{3} + \frac{12 \times 6}{12 + 6} = 6 Ω$

As given, $I_{P Q} = 0.5 A$
$∴ V_{P} - V_{Q} = I_{P Q} R_{n e t} = 0.5 A \times 6 Ω = 3.0 V$

Option C is correct.

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Resistance of 6 ohm each are connected in the manner shown in adjoining figure.With the current 0.5 ampere as shown in figure, the potential difference VP−VQ is :

The correct option is C 3.0 VFrom the figure, the net resistance between PQ is,Rnet=63+12×612+6=6ΩAs given, IPQ=0.5A∴VP−VQ=IPQRnet=0.5A×6Ω=3.0VOption C is correct.

Resistance of $6 o h m$ each are connected in the manner shown in adjoining figure.With the current $0.5$ ampere as shown in figure, the potential difference $V_{P} - V_{Q}$ is :

The correct option is C $3.0 V$
From the figure, the net resistance between PQ is,
$R_{n e t} = \frac{6}{3} + \frac{12 \times 6}{12 + 6} = 6 Ω$

As given, $I_{P Q} = 0.5 A$
$∴ V_{P} - V_{Q} = I_{P Q} R_{n e t} = 0.5 A \times 6 Ω = 3.0 V$

Option C is correct.