Engineering AC Source driving an RLC circuit....

AI Thread Summary
The discussion revolves around calculating the current and voltage across a capacitor in an RLC circuit driven by an AC source. The circuit consists of a 10V, 100Hz input with a 1Ω resistor, 1Ω inductive reactance, and 1Ω capacitive reactance in series. The correct calculation for the total impedance (Z) is 1Ω, leading to a current (I) of 10A. The voltage across the capacitor (Vc) was initially miscalculated but was corrected to -10j volts after considering the complex nature of reactance. The final consensus confirms the calculations for both current and voltage across the capacitor as accurate.
jojo13
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Homework Statement



10 v, 100 hz goes into a circuit of a 1o resistor, a 1o inductive reactance and a 1o capacitive reactance that are in series.

What is the current. What is the V across the cap.

Homework Equations

and the attempt at a solution[/B]

So I know I = V/Z and Z = sqrt( R^2 + (XL^2 - XC^2 )

And the question gives us the inductive reactance and capacitive reactance which is XL = 1 and XC = 1, and it gives us R = 1

So now, Z = sqrt(1^2 + (1^2 - 1^2 ) ) = 1

And I = V/Z = 10/1 = 10 A

Is that correct for the current of the circuit? If so, how do I calculate the voltage across the capacitor?

Is it voltage of capacitor, Vc = I * XC = 10 * 1 = 10 V ? Is that correct?
 
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Yes, that is the correct value of the current.

and the voltage across the capacitor is almost correct.

Revisit your XL and XC. Isn't there a "j" in the reactance?
 
magoo said:
Yes, that is the correct value of the current.

and the voltage across the capacitor is almost correct.

Revisit your XL and XC. Isn't there a "j" in the reactance?

Ah, would it be, Vc = I * Zc = 10 * (1/j) = 10/j = -10j
 
Now it is correct. Nice work!
 
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magoo said:
Now it is correct. Nice work!

Thanks
 
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Yes, you've got it right.
 
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