Question

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 50 μF is?

(a) 1.57 A

(b) 1.87 A

(c) 1.67 A

(d) 2.83 A

I have been asked this question in an online interview.

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

Answer 1

Correct answer is (c) 1.67 A

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

V = \(\frac{1}{C_1}  \int I_1  \,dt \)

∴ I1 = C1 \(\frac{dV}{dt}\)

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, I1 = 1.67 A.