Home  www.playhookey.com  Tue, 08042020 

Direct Current

Alternating Current

Semiconductors

Digital

Logic Families

Digital Experiments

Computers

 Analog  Analog Experiments  Oscillators  Optics  HTML Test  

 The Fundamentals  Resistance and Reactance  Filter Concepts  Power Supplies   

Series RC Circuits

Series RL Circuits

Parallel RC Circuits

Parallel RL Circuits

 Series LC Circuits  Series RLC Circuits  Parallel LC Circuits  Parallel RLC Circuits   AC Applications of the Wheatstone Bridge  
Parallel RC Circuits 

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, I_{R} is in phase with V, but I_{C} 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 I_{R} = V/R. But that +90° phase shift in C requires the use of complex numbers here, so I_{C} = j(V/X_{C}). The total current, then, is:
I =  V  =  V  + j  V 




Z  R  X_{C} 
There are two things to note about this equation. First, keep in mind that j(V/X_{C}) = V/(jX_{C}). Another way to look at this is to note that capacitive reactance is ultimately defined as X_{C} = 1/(ωC). Including the "j" factor, we have 1/(jωC). This again gives us jX_{C} 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  jX_{C} 
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(X_{L}  X_{C}) 


R + j(X_{L}  X_{C}) 
In this case, we only have R and C, so the X_{L} factors simply drop out of the equation, leaving us with:
Z  =  R × ( jX_{C}) 


R  jX_{C} 
To complete this calculation, we must remove the "j" term
from the denominator. We can do this by applying the mathematical
identity:
Z  =  R × ( jX_{C})  ×  R + jX_{C} 



R  jX_{C}  R + jX_{C}  
=  (jRX_{C})(R + jX_{C})  


(R  jX_{C})(R + jX_{C})  
=  jR²X_{C}  j²RX_{C}²  


R² + X_{C}²  
=  RX_{C}²  jR²X_{C}  


R² + X_{C}² 
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  =  RX_{C}²   j  R²X_{C} 



R² + X_{C}²  R² + X_{C}² 
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 X_{C} = 200Ω. Then:
I_{R}  =  V  =  5  =  0.05 A 



R  100  
I_{C}  =  V  =  5  =  j0.025 A 



jX_{C}  j200  
I_{T}  =  (I_{R}² + I_{C}²)^{½}  =  (0.0025 + 0.000625)^{½} = 0.0559  
Z  =  V  =  5  =  89.44Ω 



I_{T}  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 + X_{C}²  =  100² + 200²  =  10,000 + 40,000  =  50,000 
N1  =  RX_{C}²  =  100 × 200²  =  100 × 40,000  =  4,000,000 
N2  =  R²X_{C}  =  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 



D  D  
=  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.


 
All pages on www.playhookey.com copyright © 1996, 20002015 by
Ken Bigelow Please address queries and suggestions to: webmaster@playhookey.com 