Friday 7-26-96
The relevant section in the book is 20.2
Consider now a circuit which is composed of only a capacitor and an AC power source (such as a wall outlet). A capacitor is a device for storing charging. In this circuit, there will actually be a 90 degree phase difference between the current and voltage, with the current reaching its peak 90 degrees before the voltage reaches its peak. Put another way, the current leads the voltage by 90 degrees in purely capacitive circuit.
This is the reason why. The AC power supply produces an oscillating voltage. If the capacitor is the only thing in the circuit aside from the power supply, then the capacitor voltage matches the power supply voltage. Recall that for a capacitor of capacitance C, the charge stored on the capacitor is related to the voltage by Q = CV. If the voltage changes, the amount of stored charge must change, which means a current must flow in the circuit. The change in stored charge is proportional to the change in voltage; to be specific, the change in charge equals C times the change in voltage. The faster the voltage changes, the faster the stored charge changes, and the more current flows.
Consider what happens as the power supply goes through one cycle, from zero to positive to zero to negative to zero.
Step 1 - As the voltage rises the capacitor, which originally has no stored charge, starts charging. Initially, the voltage changes rapidly, so plenty of current flows. As the voltage rises it levels off more and more, so the rate of change slows, and the current drops. When the voltage reaches a peak, for an instant it is not changing at all, so the current actually falls to zero.
Step 2 - After reaching a peak, the voltage starts dropping. This causes the capacitor to start discharging, so current flows the other way. The voltage decreases slowly at first, and then drops quite quickly through zero. To reflect this, the capacitor discharges slowly (low current) and then more rapidly (high current). The current peaks when the voltage is changing most rapidly, which is when it passes through zero.
Step 3 - Now the voltage goes negative, but the capacitor is completely discharged. As the voltage decreases from zero, charge starts accumulating again on the capacitor, but it is charge opposite to the charge in step one, so the current is negative. The voltage drops from zero quite quickly, so the current is a large negative value. Gradually the current goes toward zero as the voltage change slows, the change falling to zero when the voltage reaches the negative peak.
Step 4 - The voltage rises again from its negative peak, and the capacitor starts discharging again. This discharge is opposite to the discharge in step 2, so the current is positive. The voltage rises slowly at first, so the current is low, but as the voltage rises more quickly the current increases, reaching a positive peak as the voltage passes through zero. The cycle begins again with step 1.
A capacitor in an AC circuit exhibits a kind of resistance called capacitive reactance, measured in ohms. This depends on the frequency of the AC voltage, and is given by:
Xc = 1/wC
We can use this like a resistance (because, really, it is a resistance) in an equation of the form V = IR to get the voltage across the capacitor:
V = I Xc
The Xc = 1/wC equation makes sense, because if the frequency increases, the capacitor has less time to build up the charge needed to match the voltage, so more current has to flow. Using V = IXc, more current can flow only if Xc decreases. If the frequency stayed constant but C was increased, more charge would be needed to reach the same voltage so again the current has to increase, and Xc must drop.
Note that V and I are the rms values of the voltage and current.