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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityCircuit AnalysisMedium2 minPYQ_2018

PhysicsMediumsingle choice

In the circuit shown, the current through the $5Ω$ resistor is

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Solution:

Applying Kirchhoff's second law for closed loop $A EFB A$ , we get $- (I_{1} + I_{2}) \times 5 - I_{1} \times 2 + 2 = 0$
Again, applying Kirchhoff's second law for a closed loop $D EFC D$ , we get $- (I_{1} + I_{2}) \times 5 - I_{2} \times 2 + 2 = 0$
On solving, we get $I_{1} = \frac{1}{6} A$ and $I_{2} = \frac{1}{6} A$ $I = I_{1} + I_{2} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} A$

Stream:AIIMSSubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2018

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