AC Source driving an RLC circuit....

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SUMMARY

The discussion focuses on calculating the current and voltage across a capacitor in an RLC circuit driven by a 10V AC source at 100 Hz. The circuit consists of a 1Ω resistor, 1Ω inductive reactance (XL), and 1Ω capacitive reactance (XC) in series. The impedance (Z) is calculated as 1Ω, leading to a current (I) of 10A. The voltage across the capacitor (Vc) is determined to be -10j V, correcting the initial misunderstanding regarding the use of complex numbers in reactance.

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  • Ability to apply Ohm's Law in AC circuits
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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?
 
Last edited:
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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
 
Last edited:
Yes, you've got it right.
 
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