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PhysicsCurrent ElectricityCircuit AnalysisMedium2 minPYQ_2015

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In the circuit shown, the current in the $1 Ω$ resistor is:

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Answer:

Solution:

$R = 3 Ω + 2 Ω$ . Let potential at $P = 0$ , $∴ V_{A} = 6 V, V_{D} = - 9 V$ , and $V Q = x$ . Apply K.C.L. at $x$ , $\frac{x - 6}{3} + \frac{x - 0}{1} + \frac{x + 9}{5} = 0$ . Solving, $x = \frac{3}{23} V$ . So, current in $1 Ω$ $\Rightarrow i = \frac{\frac{3}{23} - 0}{1}$ from $Q$ to $P$ $\Rightarrow i = 0.13 A$ from $Q$ to $P$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2015

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