Archived RC Circuit: Phase & Magnitude of Current

AI Thread Summary
The discussion revolves around an RC circuit analysis involving a capacitor and resistor connected to an AC voltage source. The phase of the current with respect to the applied voltage is calculated to be -27.9 degrees, indicating that the current leads the voltage. The impedance of the circuit is derived using the formulas for resistors and capacitors, resulting in a total impedance of approximately 5.66 ohms. The magnitude of the current is determined to be approximately 0.176 A, calculated by dividing the voltage by the impedance. Overall, the calculations and results presented are accurate and align with the expected outcomes for the circuit analysis.
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Homework Statement



Given an RC circuit which has a Capacitor(C=6\times 10^{-6} F) and a resistor (R=5 \Omega)conected in series to an a.c. voltage source of the form v=V_{0} e^{j\omega t}with a V_{0}=1 Volt. Frequency f=10kHz
a)What is the phase of the current with respect to the applied voltage?
b) What is the magnitude of the current?

Homework Equations



Z=Z_{R}+Z_{C}

Z_{C}=\frac{1}{j\omega C}

Z=x+jy where j=\sqrt{-1}

Z=ze^{j \phi} with z=\sqrt{x^{2}+y^{2}} and \phi =tan^{-1}\frac{y}{x}

\omega = 2\pi \times f

The Attempt at a Solution



\frac{1}{\omega C} = 2.65

Z=R+ \frac{1}{j\omega C}

=5-2.65j

Which can be written as Z=ze^{j\phi}

z=\sqrt{5^{2}+2.65^{2}}=5.66

\phi =tan^{-1}\frac{-2.65}{5}=-27.9 degrees

Since i=\frac{V_{0}}{z}e^{j(\omega t-\phi)} then I can say the current leads the voltage by 27.9 degrees.

And I can also say the magnitude of the current i is equal to \frac{V_{0}}{z}=\frac{1}{5.66}

How did I go?? Am I on track anywhere at all or have I made a bit of a mess of it?
 
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The calculations look fine. The results are good.
 
Everything look allright.
 
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