(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given an RC circuit which has a Capacitor[itex] (C=6\times 10^{-6} F) [/itex] and a resistor [itex](R=5 \Omega) [/itex]conected in series to an a.c. voltage source of the form [itex]v=V_{0} e^{j\omega t} [/itex]with a [itex]V_{0}[/itex]=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?

2. Relevant equations

[itex]Z=Z_{R}+Z_{C}[/itex]

[itex]Z_{C}[/itex]=[itex]\frac{1}{j\omega C}[/itex]

Z=x+jy where j=[itex]\sqrt{-1}[/itex]

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

[itex] \omega = 2\pi \times f[/itex]

3. The attempt at a solution

[itex] \frac{1}{\omega C} = 2.65[/itex]

Z=R+ [itex] \frac{1}{j\omega C}[/itex]

=5-2.65j

Which can be written as [itex]Z=ze^{j\phi}[/itex]

[itex]z=\sqrt{5^{2}+2.65^{2}}=5.66[/itex]

[itex]\phi =tan^{-1}\frac{-2.65}{5}=-27.9 degrees[/itex]

Since [itex]i=\frac{V_{0}}{z}e^{j(\omega t-\phi)}[/itex] 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 [itex]\frac{V_{0}}{z}=\frac{1}{5.66}[/itex]

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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# Homework Help: Archived RC circuit question

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