- #1

- 1,196

- 0

## Homework Statement

If [itex]V_{s}(t) = 12 sin(4*10^{4}t-\frac{∏}{4})[/itex], R = 2 Ω, L = .1 mH and C = .3 mF, obtain an expression for the current though the capacitor.

See attached file for circuit.

I'm really confused. I applied KVL around the loop with the coil and got that

[itex]\tilde{I} = \frac{e^{-\frac{j3∏}{4}}}{R+4*10^{4}Lj}[/itex]

I applied the KVL around the loop with the capacitor and got

[itex]\tilde{I} = \frac{4*10^{4}e^{-\frac{j3∏}{4}}}{4*10^{4}R-j}[/itex]

I don't really know where to go from here or if what I'm doing will produce the answer I'm looking for. Any help at all solving this problem would be greatly appreciated.

The first thing I did was adopt a cosine reference

[itex]V_{s}(t) = sin(4*10^{4}t-\frac{∏}{4})[/itex] = [itex]cos(4*10^{4}t-\frac{∏}{4} - \frac{∏}{2}) = cos(4*10^{4}t-\frac{3∏}{4})[/itex]

For the loop around the coil and applying KVL

[itex]cos(4*10^{4}t-\frac{3∏}{4}) = R i(t) + L \frac{di}{dt}[/itex]

Applying phasor analysis

[itex]V_{s}(t) = Re[e^{j(4*10^{4}t-\frac{3∏}{4})}] = Re[e^{j*4*10^{4}t}e^{-\frac{3∏j}{4}}] = Re[ e^{4*10^{4}tj}*\tilde{V_{s}(t)}][/itex]

And so on.

#### Attachments

Last edited: