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An a.c. source generating a voltage ε = ε0 sin ωt is connected to acapacitor of capacitance C. Find the expression for the current I flowingthrough it

Question

An a.c. source generating a voltage ε=ε0sinωt \varepsilon = \varepsilon_0 \sin \omega t is connected to a capacitor of capacitance C C . Find the expression for the current I I flowing through it.

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Solution

The current I in a capacitor connected to an AC source is given by the relationship I = C * dV/dt, where dV/dt is the rate of change of voltage.

Given the voltage ε = ε0 sin ωt, we can find dV/dt by differentiating ε with respect to time t.

d(ε0 sin ωt) / dt = ε0 * ω cos ωt

Substituting this into the equation for I gives:

I = C * ε0 * ω cos ωt

This is the expression for the current I flowing through the capacitor. The current leads the voltage by a phase of 90 degrees, which is a characteristic of capacitors in AC circuits.

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