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Capacitive CircuitsSine Wave VC Lags IC by 90For a sine wave of applied voltage, a capacitor provides a cycle of alternating charge and discharge current.

The wave shape of this charge and discharge current IC is compared with the voltage VC.

Examining the VC and IC WaveformsThe instantaneous value of IC is zero when VC is at its maximum value. At either its positive or its negative peak, VC is not changing. For one instant at both peaks, Therefore, the voltage must have a static value before changing its direction. Then V is not changing and C is not charging or discharging. The result is zero current at this time.Why IC leads VC by 90This phase between VC and IC is true in any sine-wave AC circuit, whether C is in series or parallel and whether C is alone or combined with other components. We can always say that for any XC , its current and voltage are 90 out of phase.

Capacitive Current Is the Same in a Series CircuitThe leading phase angle of capacitive current is only with respect to the voltage across the capacitor, which does not change the fact that the current is the same in all parts of a series circuit.

For instance, the current in the generator, the connecting wires, and both plates of the capacitor must be the same because they are all in the same path.

Capacitive Voltage Is the Same acrossParallel BranchesThe voltage is the same across the generator and C because they are in parallel. There cannot be any lag or lead in time between these two parallel voltages. At any instant, whatever the voltage value is across the generator at that time, the voltage across C is the same. With respect to the series current, however, both VA and VC are 90 out of phase with IC .The Frequency Is the Same for VC and ICAlthough VC lags IC by 90 , both waves have the same frequency. For example, if the frequency of the sine wave VC is 100 Hz, this is also the frequency of IC .

XC and R in SeriesWhen a capacitor and a resistor are connected in series, the current I is limited by both XC and R . The current I is the same in both XC and R since they are in series. However, each component has its own series voltage drop, equal to IR for the resistance and IXC for the capacitive reactance.

The current is labeled I , rather than IC , because I flows through all series components.The voltage across XC , labeled VC , can be considered an IXC voltage drop, just as we use VR for an IR voltage drop.The current I through XC must lead VC by 90 because this is the phase angle between the voltage and current for a capacitor.The current I through R and its IR voltage drop are in phase. There is no reactance to sine-wave alternating current in any resistance. Therefore, I and IR have a phase angle of 0.Phase ComparisonsNote the following points about a circuit containing series resistance and reactance:The voltage VC is 90 out of phase with I .However, VR and I are in phase.If I is used as the reference, VC is 90 out of phase with VR .Specifically, VC lags VR by 90 just as the voltage VC lags the current I by 90.

Phase relationships between I , VR , VC , and VT Phasors representing I , VR , and VC

Combining VR and VCWhen the voltage wave VR is combined with the voltage wave VC, the result is the voltage wave of the applied voltage VT . The voltage drops, VR and VC, must add to equal the applied voltage VT .Phasor Voltage TriangleInstead of combining waveforms that are out of phase, we can add them more quickly by using their equivalent phasors.

The phasors in (a) show the 90 phase angle without any addition.

The method in (b) is to add the tail of one phasor to the arrowhead of the other, using theangle required to show their relative phase.(a)(b)Impedance Z TriangleA triangle of R and XC in series corresponds to the voltage triangle. It is similar to the voltage triangle, but the common factor I cancels because the series currentI is the same in XC and R

Phase Angle with Series XC and RThe angle between the applied voltage VT and the series current I is the Phase Angle of the circuit. Its symbol is (theta). In the corresponding impedance triangle, the angle between Z T and R is also equal to the phase angle. Therefore, the phase angle can be calculated from the impedance triangle of a series RC circuit by the formula

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Sample:

Series Combinations of XC and RIn series, the higher the XC compared with R , the more capacitive the circuit. There is more voltage drop across the capacitive reactance XC , and the phase angle increases toward 90. The series XC always makes the series current I lead the applied voltage VT . With all XC and no R , the entire applied voltage VT is across XC and equals 90.

RC Phase-Shifter CircuitApplication of XC and R in series to provide a desired phase shift in the output V R compared with the input V T . The R can be varied up to 100 k to change the phase angle. The C is 0.05 F here for the 60-Hz ac power-line voltage, but a smaller C would be used for a higher frequency. The capacitor must have an appreciable value of reactance for the phase shift.

For the circuit in Fig. (a), assume that R is set for 50 k at its middle value. The reactance of the 0.05-F capacitor at 60 Hz is approximately 53 k. For these values of XC and R , the phase angle of the circuit is 46.7. This angle has a tangent of 5350 1.06.The Phasor Triangle in Fig. (b) shows that IR or VR is out of phase with VT by the leading angle of 46.7. Note that VC is always 90 lagging VR in a series circuit.The angle between VC and VT then becomes 90-46.7= 43.3

XC and R in ParallelFor parallel circuits with XC and R , the 90 phase angle must be considered for each of the branch currents. Remember that any series circuit has different voltage drops but one common current. A parallel circuit has different branch currents but one common voltage.

In the parallel circuit in Fig. (a) , the applied voltage V A is the same across X C , R , and the generator, since they are all in parallel. There cannot be any phase difference between these voltages. Each branch, however, has its own individual current.For the resistive branch, IR VA R; for the capacitive branch, IC VC XC .The resistive branch current I R is in phase with the generator voltage V A . The capacitive branch current IC leads VA , however, because the charge and dischargeFor the circuit in Fig. (a), VA is 100 V, and the total current IT , obtained as the phasor sum of IR and IC , is 14.14 A. Therefore, we can calculate the equivalent impedance ZEQ asImpedance of XC and R in Parallel

A practical approach to the problem of calculating the total or equivalent impedance of XC and R in parallel is to calculate the total line current IT and divide the applied voltage VA by this value.

(a)

Phase Angle in Parallel CircuitsIn this figure, the phase angle is 45 because R and XC are equal, resulting in equal branch currents. The phase angle is between the total current IT and the generator voltage VA . However, VA and IR are in phase. Therefore is also between IT and IR .

Using the tangent formula to find from the current triangle in Figure gives:

Parallel Combinations of XC and RWhen XC is 10 times R , the parallel circuit is practically resistive because there is little leading capacitive current in the main line. The small value of IC results from the high reactance of shunt XC . Then the total impedance of the parallel circuit is approximately equal to the resistance, since the high value of XC in a parallel branch has little effect. The phase angle of 5.7 is practically 0 because almost all of the line current is resistive

As XC becomes smaller, it provides more leading capacitive current in the main line. When XC is 110 R , practically all of the line current is the IC component. Then, the parallel circuit is practically all capacitive with a total impedance practically equal to XC . The phase angle of 84.3 is almost 90 because the line current is mostly capacitive. Note that these conditions are opposite to the case of XC and R in series. With XC and R equal, their branch currents are equal and the phase angle is 45.