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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityCircuit AnalysisEasy2 minPYQ_2020

PhysicsEasysingle choice

In the adjoining circuit diagram, $E = 5 V$ , $r = 1 Ω$ , $R_{2} = 4 Ω$ , $R_{1} = R_{3} = 1 Ω$ and $C = 3 mF$ . Then the numerical value of the charge on each plate of the capacitor is

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Answer:

Solution:

Current through resistors $R_{1}$ and $R_{3}$ in steady state is zero.

Therefore current through $R_{2}$ is

$i_{2} = \frac{E}{R_{2} + r} = \frac{5}{4 + 1} = 1 A$

Therefore charge on capacitor is

$q = C (\frac{i_{2} R_{2}}{2}) = 6 μC$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2020

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