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Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductive reactance impedes its flow. With AC, there is no time for the current to become extremely large.

Capacitors and capacitive reactance

Consider the capacitor connected directly to an AC voltage source as shown in [link] . The resistance of a circuit like this can be made so small that it has a negligible effect compared with the capacitor, and so we can assume negligible resistance. Voltage across the capacitor and current are graphed as functions of time in the figure.

Part a of the figure shows a capacitor C connected across an A C voltage source V. The voltage across the capacitor is given by V C. Part b of the diagram shows a graph for the variation of current and voltage across the capacitor as functions of time. The voltage V C and current I C is plotted along the Y axis and the time t is along the X axis. The graph for current is a progressive sine wave from the origin starting with a wave along the negative Y axis. The graph for voltage is a cosine wave and amplitude slightly less than the current wave.
(a) An AC voltage source in series with a capacitor C having negligible resistance. (b) Graph of current and voltage across the capacitor as functions of time.

The graph in [link] starts with voltage across the capacitor at a maximum. The current is zero at this point, because the capacitor is fully charged and halts the flow. Then voltage drops and the current becomes negative as the capacitor discharges. At point a, the capacitor has fully discharged ( Q = 0 size 12{Q=0} {} on it) and the voltage across it is zero. The current remains negative between points a and b, causing the voltage on the capacitor to reverse. This is complete at point b, where the current is zero and the voltage has its most negative value. The current becomes positive after point b, neutralizing the charge on the capacitor and bringing the voltage to zero at point c, which allows the current to reach its maximum. Between points c and d, the current drops to zero as the voltage rises to its peak, and the process starts to repeat. Throughout the cycle, the voltage follows what the current is doing by one-fourth of a cycle:

Ac voltage in a capacitor

When a sinusoidal voltage is applied to a capacitor, the voltage follows the current by one-fourth of a cycle, or by a 90º phase angle.

The capacitor is affecting the current, having the ability to stop it altogether when fully charged. Since an AC voltage is applied, there is an rms current, but it is limited by the capacitor. This is considered to be an effective resistance of the capacitor to AC, and so the rms current I size 12{I} {} in the circuit containing only a capacitor C size 12{C} {} is given by another version of Ohm’s law to be

I = V X C , size 12{I= { {V} over {X rSub { size 8{C} } } } } {}

where V size 12{V} {} is the rms voltage and X C size 12{X rSub { size 8{C} } } {} is defined (As with X L size 12{X rSub { size 8{L} } } {} , this expression for X C size 12{X rSub { size 8{C} } } {} results from an analysis of the circuit using Kirchhoff’s rules and calculus) to be

X C = 1 fC , size 12{X rSub { size 8{C} } = { {1} over {2π ital "fC"} } } {}

where X C size 12{X rSub { size 8{C} } } {} is called the capacitive reactance    , because the capacitor reacts to impede the current. X C size 12{X rSub { size 8{C} } } {} has units of ohms (verification left as an exercise for the reader). X C size 12{X rSub { size 8{C} } } {} is inversely proportional to the capacitance C size 12{C} {} ; the larger the capacitor, the greater the charge it can store and the greater the current that can flow. It is also inversely proportional to the frequency f size 12{f} {} ; the greater the frequency, the less time there is to fully charge the capacitor, and so it impedes current less.

Calculating capacitive reactance and then current

(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current if the applied rms voltage is 120 V?

Strategy

The capacitive reactance is found directly from the expression in X C = 1 fC size 12{X rSub { size 8{C} } = { {1} over {2π ital "fC"} } } {} . Once X C has been found at each frequency, Ohm’s law stated as I = V / X C size 12{I=V/X rSub { size 8{C} } } {} can be used to find the current at each frequency.

Solution for (a)

Entering the frequency and capacitance into X C = 1 fC size 12{X rSub { size 8{C} } = { {1} over {2π ital "fC"} } } {} gives

X C = 1 fC = 1 6 . 28 ( 60 . 0 / s ) ( 5 . 00  μ F ) = 531 Ω at 60 Hz . alignl { stack { size 12{X rSub { size 8{C} } = { {1} over {2π ital "fC"} } } {} #" "= { {1} over {6 "." "28" \( "60" "." 0/s \) \( 5 "." "00" μF \) } } ="531 " %OMEGA " at 60 Hz" {} } } {}

Similarly, at 10 kHz,

X C = 1 fC = 1 6 . 28 ( 1 . 00 × 10 4 / s ) ( 5 . 00  μ F ) = 3.18 Ω at 10 kHz . alignl { stack { size 12{X rSub { size 8{C} } = { {1} over {2π ital "fC"} } = { {1} over {6 "." "28" \( 1 "." "00" times "10" rSup { size 8{4} } /s \) \( 5 "." "00" μF \) } } } {} #" "=3 "." "18" %OMEGA " at 10 kHz" {} } } {}

Solution for (b)

The rms current is now found using the version of Ohm’s law in I = V / X C size 12{I=V/X rSub { size 8{C} } } {} , given the applied rms voltage is 120 V. For the first frequency, this yields

I = V X C = 120 V 531 Ω = 0.226 A at 60 Hz . size 12{I= { {V} over {X rSub { size 8{C} } } } = { {"120"" V"} over {"531 " %OMEGA } } =0 "." "226"" A"} {}

Similarly, at 10 kHz,

I = V X C = 120 V 3.18 Ω = 37.7 A at 10 kHz . size 12{I= { {V} over {X rSub { size 8{C} } } } = { {"120"" V"} over {3 "." "18 " %OMEGA } } ="37" "." 7" A"} {}

Discussion

The capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. At the higher frequency, its reactance is small and the current is large. Capacitors favor change, whereas inductors oppose change. Capacitors impede low frequencies the most, since low frequency allows them time to become charged and stop the current. Capacitors can be used to filter out low frequencies. For example, a capacitor in series with a sound reproduction system rids it of the 60 Hz hum.

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Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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