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Question
An AC source generating a voltage e = e0sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus ωt.
Solution
The following figure shows an AC source, generating a voltage e = e0 sin wt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the
An AC source connected to a capacitor
capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = `"q"/"C"` ∴ q = CV. q and V are functions of time, with V = e = e0 sin ωt.
The instantaneous current in the circuit is i = `"dq"/"dt" = "d"/"dt" ("CV") = "C" "dV"/"dt"`
`= "C" "d"/"dt" ("e"_0 sin omega"t") = omega "C""e"_0 cos omega"t"`
∴ i = `"e"_0/((1//omega"C")) sin (omega"t" + pi/2) = "i"_0 sin (omega"t" + pi/2)`
where `"i"_0 = "e"_0/((1//omega"C"))` is the peak value of the current.
| ωt (rad) | ωt + π/2 (rad) | e = e0 sin ωt | i = i0 sin `(omega"t" + pi/2)` |
| 0 | `pi/2` | 0 | i0 |
| `pi/2` | π | e0 | 0 |
| π | `(3pi)/2` | 0 | - i0 |
| `(3pi)/2` | 2π | - e0 | 0 |
| 2π | 2π + `pi/2` | 0 | i0 |
The above table shows gives the values of e and i for different values of ωt and the following figure shows graphs of e and i versus ωt. i leads e by a phase angle of π/2 rad.
Graphs of e and i venus ωt for a purely capacitive AC circuit
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