An AC source generating a voltage e = e_{0}sinωt 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 = e_{0} 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 = e_{0} 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 = e_{0} sin ωt |
i = i_{0} sin `(omega"t" + pi/2)` |

0 | `pi/2` | 0 | i_{0} |

`pi/2` | π | e_{0} |
0 |

π | `(3pi)/2` | 0 | - i_{0} |

`(3pi)/2` | 2π | - e_{0} |
0 |

2π | 2π + `pi/2` | 0 | i_{0} |

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 t and i venus wt for a purely capacitive AC circuit**