The R-C circuit contains resistors and capacitors, as shown in Figure 1. The unit of capacitance (C) is farad. The units of resistance (R) and impedance (Z) are both ohms. But how to combine R and C in different units into impedance Z to get the total resistance to the current? AC circuits containing capacitors or inductors are called reactance circuits. The inductance Xc is the resistance of the capacitor to the current, in ohms. Therefore, the units of Xc, R and Z are all ohms. Capacitive reactance is a function of frequency and capacitance, defined as Xc=1/2πfC. In the formula: frequency is f; capacitance is C; π=3.14. Xc is not considered in the DC circuit, because the frequency of DC is zero. Please note that when the capacitance value is constant, Xc is controlled by frequency changes.

**1. Calculation of R-C series circuit**When analyzing AC reactance circuits, the impedance (Z) formula can be used to combine the data. The reason for using this formula to calculate impedance is that the phase difference between resistance and capacitive reactance is 90°. In the series R-C circuit, the total resistance of the current can be calculated by the formula Z=√(R²+X²). If the resistance of the circuit is 6Ω and Xc is 8Ω, then Z=√(6²+8²)Ω=√(36+64)Ω=10Ω. Ohm’s law can be applied to the RC circuit, but when calculating the total voltage (ET) or total For current (IT), replace resistance (R) with impedance (Z). So we get E=I×Z, and the impedance is equal to the total voltage (ET) divided by the total current (IT), that is, Z=ET/IT.

The DC circuit is a pure resistance circuit, so the voltage and current are in phase. When a capacitor is added to the circuit and the power supply becomes AC, there will be a phase difference between the power supply voltage and current.

example 1

Use the values in Figure 2 to calculate Z, I, V_{R}, and V_{C}.

(1) Calculate Z.

Z=√(R²+X²)=√(2500²+4000²)Ω

=√(6250000+1667089)Ω =4717Ω

(2) I is determined by Ohm’s law

I=E/Z=120/4717A=0.0254A=25.4mA

(3) V_{R} and V_{C} are

V_{R}=I×R=0.0254×2500V=63.5V

V_{C}=I×Xc=0.0254×4000V=101.6V

(4) Verify that the total voltage drop is equal to the power supply voltage:

E=√(V²_{R}+V²_{C})=√(63.5²+101.6²)V=119.8V

Example 2

A wind turbine drives a load (R_{L)}, and the load is connected in series with a capacitor and a resistor. Use the values in Figure 3 to calculate Z, I, V_{R}, V_{Xc}, and V_{RL}. First consider the case with Xc, and then consider the case without Xc.

(1) When Xc is present:

Z=√(R²+X²_{C})=√[(0.5+6)²+3²]Ω=√(6.5²+3²)Ω=7.15Ω

Apply Ohm’s Law:

I=E/Z=120/7.15A=16.8A

Vg =I×R=16.8×0.5V=8.4V

VXc=I×Xc=16.8 ×3V=50.4V

VRL=I×RL=16.8×6V= 100.8V

Use the following formula to verify E

E=√(V²R+V²Xc)

=√(8.4 + 100.8)²+50.4v

=120V

(When the local pressure drops several times. Follow this method.)

(2) When there is no Xc:

Z=R_{T}=R+R_{L}=0.5 +6=6.5Ω (remember that the resistance R can be added directly)

I_{T}=E_{T}/R=120/6.5 =18.46A

V_{R}=I×R=18.46 x0.5=9.2V

V_{RL}=I×R_{L}=18.46 x6=110.8V

Verify that the sum of pressure drops is equal to E:

V_{R }+ V_{RL}=9.2 +110.8 = 120V

(Remember that the pressure drop across R can be added directly.)

This example shows the influence of Xc on the circuit and the analysis and calculation process after adding resistance and reactance to a purely resistive circuit.

**2. Calculation of R-C parallel circuit**

All resistance to the current can be carried out with the formula R_{T}=(R_{1}×R_{2})/(R_{1}+R_{2})calculate. But in the AC R-C parallel circuit, the total impedance is Z=(R×Xc)/√(R²+X²C). If R=6Ω and Xc=8Ω, then Z=(6×8)/√(6²-+8²)Ω=48/10Ω=4.8Ω.

Example 3

Using the values in Figure 4, calculate Z, V_{R}, V_{Xc}, I_{R} and I_{Xc}.

Impedance is

Z=(R×Xc)/√(R²+X²)

=(2500×4000)/√(2500²+4000²)Ω=2120Ω

The voltage is

V_{R}=V_{Xc}=24V (remember, the voltage drop of each parallel branch is the same)

The current is

I_{R}=V_{R}/R=24/1400A=9.6mA

I_{Xc}=V_{Xc}/Xc=24/4000A =6mA

Example 4

Use the values in Figure 5 to calculate I_{R}, I_{Xc}, and I_{T}. Among them, the voltage V_{R}=Vc=E.

(1) Calculate current:

I_{R} =V_{R}/R=80/1000A=80mA

I_{Xc}=V_{Xc}/Xc=80/2123A=37.7mA

(3) Calculate I_{T}

I_{T}=√(I²_{R}+I²_{Xc})=√(80mA)²+(37.7mA)²=88.4mA

(When increasing the number of branch currents, follow this method.)

Let us recall that in a purely resistive parallel circuit, the branch current can be directly accumulated (I_{1}+I_{2}+I_{3}+…+I_{n}) to get the total current. But in the reactance circuit, the total current (I_{T}) is the square root of the sum of the squares of the currents of each branch, as shown in this example.