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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 AThis 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

Answer»

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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