Rlc circuit with ac current source

AI Thread Summary
An AC current source of 1 ∠0° A at 1 rad/s is applied to an RLC circuit consisting of a 1F capacitor in parallel with a 1H inductor and a 1Ω resistor in series. The voltage across the capacitor is equal to the voltage into the circuit due to their parallel arrangement. To find this voltage, the current through the capacitor must be determined using the current divider rule and the impedances of the components. The impedance of the capacitor is calculated as 1/(ωC) ∠-90°, while the total impedance of the inductor and resistor needs to be combined. Understanding complex impedance is crucial for solving this problem effectively.
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Homework Statement


An AC current source, 1 ∠0° A, with angular frequency 1 rad/s is to be applied to the input
terminals of the circuit. Circuit is RLC, the current source is in parallel with a 1F capacitor which is in parallel with a 1H inductor which is in series with a 1Ω resistor. I need to find the Voltage across the capacitor.


Homework Equations



Xc = 1/ωC

Vc = Xc * I



The Attempt at a Solution



I know that the voltage into the circuit is the same as the voltage across the capacitor since they are in parallel. This is a current divider problem so I need to find the current across the capacitor to solve for the Vc. I know I need to use the impedences to solve for voltage using Vc = Xc * Ic, but I cannot recall how to do this for a series/parallel RLC circuit.
 
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Is it that impedences act the same as resistors? In that case I just add the series elements and then use the current divider rule to find the current across the capacitor, but I don't know if that is right.
 
First find the net impedence of the circuit.
 
Have you been introduced to the concept of complex imedance?
 
Step 1. The impedance of the capacitor = 1/(wC) ∠-90°

Step 2. Impedance of the inductor + impedance of resistor = [PLAIN][PLAIN]https://www.physicsforums.com/images/icons/icon5.gif ∠[PLAIN][PLAIN]https://www.physicsforums.com/images/icons/icon5.gif  + [PLAIN][PLAIN]https://www.physicsforums.com/images/icons/icon5.gif ∠[PLAIN][PLAIN]https://www.physicsforums.com/images/icons/icon5.gif
 
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