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Parallel RC Circuits

The Circuit

A parallel RC circuit.

The parallel RC circuit shown to the right behaves very differently when AC is applied to it, than when DC is applied. With a DC voltage, the capacitor will charge rapidly to that voltage, after which the only current flowing will be through the resistor. But with an applied AC voltage, the capacitor cannot ever reach a final charge, and therefore will always be carrying some current.

To make matters more interesting, we know that the voltage in a parallel circuit must be the same throughout the circuit. However, the current through R is not the same as the current through C. Thus, IR is in phase with V, but IC leads V by 90°.

Because the voltage is everywhere the same in this circuit, we must use voltage as the reference, and determine the total circuit current in terms of that voltage. To do this, we turn to Ohm's Law. We already know that IR = V/R. But that +90° phase shift in C requires the use of complex numbers here, so IC = j(V/XC). The total current, then, is:

I =  V  =  V  + j V


There are two things to note about this equation. First, keep in mind that j(V/XC) = V/(-jXC). Another way to look at this is to note that capacitive reactance is ultimately defined as XC = 1/(ωC). Including the "j" factor, we have 1/(jωC). This again gives us -jXC for capacitive reactance.

The second point is, since V is the same in all terms of this equation, we can divide each term by V and thus remove it from our calculations. Therefore, the complex equation we really need to solve is:

1  =  1  +  1

Z R -jXC

The Complex Number Calculations

The above equation is simply the beginning of the general equation for impedances in parallel. The only twist is that one is a reactance while the other is a resistance. Therefore, we need to deal with that pesky "j" in one term. Fortunately, this isn't as difficult as it might seem at first.

The expression for impedances in parallel is simply an update of the expression for resistors in parallel. If we first collect any multiple resistors together to form a composite R, and also determine a single composite C and a single composite L, the general expression for these elements in parallel becomes:

Z  =  R × j(XL - XC)

R + j(XL - XC)

In this case, we only have R and C, so the XL factors simply drop out of the equation, leaving us with:

Z  =  R × (- jXC)

R - jXC

To complete this calculation, we must remove the "j" term from the denominator. We can do this by applying the mathematical identity:

(a + b)(a - b) = a² - b²

In this case,
(R + jXC)(R - jXC) = R² - j²XC² = R² + XC²

Thus, we can multiply the parallel expression by (R+jXC)/(R+jXC) and get the following result:

Z  =  R × (- jXC)  ×  R + jXC

R - jXC R + jXC
   =  (-jRXC)(R + jXC)

(R - jXC)(R + jXC)
   =  -jR²XC - j²RXC²  

R² + XC²  
   =  RXC² - jR²XC  

R² + XC²  

This expression gives us an entirely real number in the demoninator, which in turn makes the necessary computations possible and practical. Our parallel RC impedance now has a real term and a "j" term, and can be written as:

Z  =  RXC²  - j R²XC

R² + XC² R² + XC²

This expression can now be used to calculate the parallel impedance of any resistor and any capacitor, provided the signal frequency is known.

To verify this mathematical expression, let's try a practical example. Let V = 5 volts RMS, with R = 100Ω and XC = 200Ω. Then:

IR  =  V  =  5  =  0.05 A

R 100
IC  =  V  =  5  =  j0.025 A

-jXC -j200
IT  =  (IR² + IC²)½  =  (0.0025 + 0.000625)½ = 0.0559
Z  =  V  =  5  =  89.44Ω

IT 0.0559

The next step is to calculate Z using the equation we derived earlier, and compare that result with the result above. If we've done our math correctly, the results should match. For simplicity we will first calculate the denominator (D) value and the two numerators (N1 and N2). Then we can insert those values into the final equation.

D  =  R² + XC²  =  100² + 200²  =  10,000 + 40,000  =  50,000
N1  =  RXC²  =  100 × 200²  =  100 × 40,000  =  4,000,000
N2  =  R²XC  =  100² × 200  =  10,000 × 200  =  2,000,000

Now we can insert these values into the full equation and solve for Z:

Z  =  N1  - j N2

   =  4,000,000  - j 2,000,000

50,000 50,000
   =  80  - j 40
Z  =  (80² + 40²)½  =  (8000)½
   =  89.44Ω    

We see that both sets of calculations produce precisely the same answer. This indiates that our method for calculating impedance without using (or knowing) the signal voltage is perfectly valid.

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