Engineering RLC circuit -- determine the voltage across each element

AI Thread Summary
In the series RLC circuit discussion, participants calculated the total impedance, current, and voltages across each circuit element. The total impedance was found to be 82.76∠-25.02, leading to a current of 0.12∠25.02. The voltage across the resistor was confirmed as 9V, while the voltages across the capacitor and inductor were calculated as 7.2∠-64.98 and 3∠115.02, respectively. It was emphasized that the voltages must be added as complex numbers to ensure they equal the source voltage of 10V. The discussion also touched on how the calculations would differ if the components were arranged in parallel.
TheRedDevil18
Messages
406
Reaction score
2

Homework Statement


For the series RLC circuit shown in Fig. Q8, determine the voltage across each element, and draw a complete phasor diagram.

os69eg.jpg


Homework Equations

The Attempt at a Solution



Total Impedance:
Z = R+Xc+Xl
= 75 - 60j + 25j
= 75-j35

Z = 82.76∠-25.02 (Phasor form)

Total Current:
I = V/Z
= (10∠0)/(82.76∠-25.02)
= 0.12∠25.02

Now my question is when finding the voltage across the resistor will it just be the magnitude of the current*resistance ie, 0.12*75 ?

Also when finding the voltage across the inductor, is this correct ?
V = (0.12∠25.02)*(75.06∠18.43)......Z = 75+j25, in phasor form = 75.06∠18.43
= 9∠43.45

And the capacitor
V = (0.12∠25.02)*(96.05∠-38.66)......Z = 75-j60, in phasor form = 96.05∠-38.66
= 11.53∠-13.64

Thanks
 
Physics news on Phys.org
TheRedDevil18 said:
Now my question is when finding the voltage across the resistor will it just be the magnitude of the current*resistance ie, 0.12*75 ?
Sure.

If you add all values you should get the 10 V of the source back. Does that happen?
 
mfb said:
Sure.

If you add all values you should get the 10 V of the source back. Does that happen?

Okay, so the voltage across the resistor would be V = 0.12*75 = 9V

For the voltage across the capacitor, it should be
V = I*Z
= (0.12∠25.02)*(60∠-90)
= 7.2∠-64.98

Voltage across the inductor
V = I*Z
= (0.12∠25.02)*(25∠90)
= 3∠115.02

But 9+7.2+3 does not equal the voltage of the source ?
 
You have to take the angles into account. Or calculate the real and imaginary part separately.
 
TheRedDevil18 said:
Okay, so the voltage across the resistor would be V = 0.12*75 = 9V
That will give you the magnitude of the voltage across the resistor, but won't give you the phase of that voltage. Use the complex current for the calculation.
For the voltage across the capacitor, it should be
V = I*Z
= (0.12∠25.02)*(60∠-90)
= 7.2∠-64.98

Voltage across the inductor
V = I*Z
= (0.12∠25.02)*(25∠90)
= 3∠115.02

But 9+7.2+3 does not equal the voltage of the source ?
The voltages are all complex values. Add appropriately.
 
Okay,

Voltage of resistor
Vr = 9∠25.02

Voltage of inductor
Vl = 3∠115.02

Voltage of capacitor
Vc = 7.2∠-64.98

So, Vt = Vr+Vl+Vc
= 3.05-6.52j-1.27+2.72j+8.16+3.81j
Vt = 9.94+0.01j

|Vt| = 9.94V
 
Looks good.

Keep a few extra digits in intermediate values in order to prevent truncation and roundoff errors from creeping into final values. Round results for presentation to the required sig figs after you're done calculating.
 
Ok, thanks guys

One last question, if they where all in parallel then the total impedance would be,
1/Z = 1/R + 1/Xc + 1/Xl ?
 
TheRedDevil18 said:
Ok, thanks guys

One last question, if they where all in parallel then the total impedance would be,
1/Z = 1/R + 1/Xc + 1/Xl ?
Yes.
 

Similar threads

Replies
3
Views
4K
Replies
13
Views
36K
Replies
2
Views
2K
Back
Top