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A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 100 μF is?

(a) 2.33 A

(b) 3.33 A

(c) 1.33 A

(d) 4.33 A

This question was posed to me in semester exam.

The query is from Problems Involving Dot Conventions topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory

1 Answer

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The correct option is (b) 3.33 A

Explanation: As the capacitors are in parallel, then the voltage V is given by,

V =  \(\frac{1}{C_2} \int I_2  \,dt \)

∴ I2 = C2 \(\frac{dV}{dt}\)

That is, \(\frac{I_1}{I_2}  = \frac{C_1}{C_2}  = \frac{50}{100} = \frac{1}{2}\) ………………….. (1)

Also, I1 + I2 = 5 A ………………………….. (2)

Solving (1) and (2), we get, I2 = 3.33 A.

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