In the circuit, the potential difference across PQ will be nearest to

9.6 V

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6.6 V

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4.8 V

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3.2 V

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Solution

The correct option is D 3.2 V
Potential difference across PQ i.e. p.d across the resistance of 20 $Ω$ , which is V= $i \times$ 20
and $i = \frac{48}{(100 + 100 + 80 + 20)} = 0.16 A$
$∴$ V=0.16 $\times$ 20=3.2V.

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In the circuit, the potential difference across PQ will be nearest to

The correct option is D 3.2 VPotential difference across PQ i.e. p.d across the resistance of 20Ω, which is V=i×20 and i=48(100+100+80+20)=0.16 A ∴ V=0.16×20=3.2V.

The correct option is D 3.2 V
Potential difference across PQ i.e. p.d across the resistance of 20 $Ω$ , which is V= $i \times$ 20
and $i = \frac{48}{(100 + 100 + 80 + 20)} = 0.16 A$
$∴$ V=0.16 $\times$ 20=3.2V.