How to Calculate Impedance in an AC Circuit with Series Components?

AI Thread Summary
To calculate impedance in an AC circuit with a 25-Ω resistor, a 30-mH inductor, and a 12-µF capacitor connected to a 90-V (rms) source at 500 Hz, one must first determine the inductive and capacitive reactances using the formulas X_L = ωL and X_C = 1/(ωC), where ω is the angular frequency. The total impedance Z can be expressed as Z = R + i(X_L - X_C), incorporating both resistance and reactance. The phase angle can be calculated from the impedance, and the current in the circuit can be found using Ohm's law. Finally, the voltage across each component and the power dissipated in the circuit can be derived from the calculated values. Understanding these calculations is essential for analyzing AC circuits effectively.
9.8m*s^-2
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Homework Statement



A 25-Ω resistance is connected in series with a 30-mH inductance and a 12-µF capacitor and are connected to a 90-V (rms) ac. If the frequency of the ac source is 500 Hz, calculate

(i) the current in the circuit.
(ii) the voltage across each element
(iii) the phase angle
(iv) the power dissipated in the circuit



Homework Equations





The Attempt at a Solution



I believe I have to begin this problem by calculating the impedance but I do not understand how to begin doing that.

Thank you.
 
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X_L=\omega L
X_C=\frac{1}{\omega C}
Z= R + i(X_L - X_C)
 
Well Impedence for each element is given by the equations that xcvxcvvc gave except with one little adjustment:

Z_L=j\omega L
Z_C=\frac{1}{j\omega C}
(The ones he gave were reactance)

\omega is the angular frequency which can be calculated by setting it equal to 2\pi f

From there you can solve the circuit.
 
Also, just in case you don't see it:
j = \sqrt{-1}
\frac{1}{j}= \frac{j}{j}\frac{1}{j} =\frac{j}{-1}=-j

which is why that capacitive reactance is subtracted when forming the expression for impedance.
 
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