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Hello there, I'm not sure if my solution is correct for [itex]\hat{I}_{C}[/itex]

In the given circuit, calculate the current in each circuit element given that V = V[itex]_{o}[/itex]sin(ωt)

R, L and C are given.

http://imgur.com/yO3flg8

Z = R + jX (j[itex]^{2}[/itex] = -1)

X[itex]_{L}[/itex] = jωL

X[itex]_{C}[/itex] = [itex]\frac{-j}{ωC}[/itex]

[itex]\hat{V}[/itex] = V[itex]_{o}[/itex]e[itex]^{jωt}[/itex]

[itex]\hat{I}[/itex] = I[itex]_{o}[/itex]e[itex]^{j(ωt-ø)}[/itex]

I[itex]_{o}[/itex] = [itex]\frac{V_{}}{|Z|}[/itex]

tan(ø) = [itex]\frac{\Im(Z)}{\Re(Z)}[/itex]

First for the impedance, [itex]1/Z_{XL} = 1/X_{L} + 1/X_{C}[/itex]

[itex] Z_{XL} = j\frac{ωL}{1- ω^{2}CL} [/itex]

[itex] Z = R + Z_{XL} = R + j\frac{ωL}{1- ω^{2}CL} [/itex]

The current phasor in the resistor R: [itex]\hat{I_{R}}[/itex]= [itex]\hat{I}[/itex]= [itex]\hat{I_{C}} + \hat{I_{L}}[/itex]

[itex]\hat{V_{C}} = \hat{V} - \hat{V_{R}} [/itex]

[itex]\hat{I_{C}} = (\hat{V} - \hat{V_{R}} )/X_{C}[/itex]

[itex]\hat{I_{C}} = (V_{o}e^{jωt} - I_{o}e^{j(ωt-ø)})(jωc) = V_{o}ωCe^{jwt}(1-\frac{R}{|Z|}e^{-jø})(j) [/itex]

[itex]\hat{I_{C}} = V_{o}ωCe^{j(ωt+\pi/2)}(1-\frac{R}{|Z|}e^{-jø})[/itex] ø and |Z| are known.

And[itex]I_{L}[/itex] can be found the same way.

I'm not entirely sure my solution is correct. Also, since [itex]I_{R}[/itex] is always in phase with V, does that mean that ø is 0? If so, then tan(ø) is also 0 but that would mean that either ω or L are 0...

Note: sorry if it seems slobby and for the skipped steps, my exam is in less than 3 hours and I'm really nervous and running out of time.

## Homework Statement

In the given circuit, calculate the current in each circuit element given that V = V[itex]_{o}[/itex]sin(ωt)

R, L and C are given.

http://imgur.com/yO3flg8

## Homework Equations

Z = R + jX (j[itex]^{2}[/itex] = -1)

X[itex]_{L}[/itex] = jωL

X[itex]_{C}[/itex] = [itex]\frac{-j}{ωC}[/itex]

[itex]\hat{V}[/itex] = V[itex]_{o}[/itex]e[itex]^{jωt}[/itex]

[itex]\hat{I}[/itex] = I[itex]_{o}[/itex]e[itex]^{j(ωt-ø)}[/itex]

I[itex]_{o}[/itex] = [itex]\frac{V_{}}{|Z|}[/itex]

tan(ø) = [itex]\frac{\Im(Z)}{\Re(Z)}[/itex]

## The Attempt at a Solution

First for the impedance, [itex]1/Z_{XL} = 1/X_{L} + 1/X_{C}[/itex]

[itex] Z_{XL} = j\frac{ωL}{1- ω^{2}CL} [/itex]

[itex] Z = R + Z_{XL} = R + j\frac{ωL}{1- ω^{2}CL} [/itex]

The current phasor in the resistor R: [itex]\hat{I_{R}}[/itex]= [itex]\hat{I}[/itex]= [itex]\hat{I_{C}} + \hat{I_{L}}[/itex]

[itex]\hat{V_{C}} = \hat{V} - \hat{V_{R}} [/itex]

[itex]\hat{I_{C}} = (\hat{V} - \hat{V_{R}} )/X_{C}[/itex]

[itex]\hat{I_{C}} = (V_{o}e^{jωt} - I_{o}e^{j(ωt-ø)})(jωc) = V_{o}ωCe^{jwt}(1-\frac{R}{|Z|}e^{-jø})(j) [/itex]

[itex]\hat{I_{C}} = V_{o}ωCe^{j(ωt+\pi/2)}(1-\frac{R}{|Z|}e^{-jø})[/itex] ø and |Z| are known.

And[itex]I_{L}[/itex] can be found the same way.

I'm not entirely sure my solution is correct. Also, since [itex]I_{R}[/itex] is always in phase with V, does that mean that ø is 0? If so, then tan(ø) is also 0 but that would mean that either ω or L are 0...

Note: sorry if it seems slobby and for the skipped steps, my exam is in less than 3 hours and I'm really nervous and running out of time.

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