Complex numbers simple circuit

[elcotufa]

Homework Statement

 __Is_____R_____
 | 		|
 |             |
+|	       ___ C	
 Vs		_
-|		|
 |		|             
 |_____________|

Find Is

The Attempt at a Solution

$$Z= R+\frac1{jwC}$$

$$Vs=coswt$$

$$Is=\frac{coswt}{R+\frac1{jwC}} * \frac{(jwC)}{(jwC)}$$

$$Is=\frac{coswt*jwC}{RjwC+1} * \frac{(1-RjwC)}{(1-RjwC)}$$

$$Is=\frac{coswt*jwC+coswt*R(wC)^2}{1+(RwC)^2}$$

$$Is=\frac{coswt*wC(j+RwC)}{1+(RwC)^2}$$



My book is pretty crappy so there are no examples

Is it right until that point? Can I simplify it some more? what is the best form to put it on? how would it look in exponential form?



Input appreciated

 

[AvroArrow]

! Yes, your solution is correct until that point. To simplify it further, you could write it in exponential form: Is = cos(wt) * wC * e^(j*(wt+RwC)) / (1+(RwC)^2).

 

[baba89]

.





Your attempt at finding Is is correct until the final step. To simplify it further, you can use the trigonometric identity coswt = (e^(jwt) + e^(-jwt))/2. This will give you:

Is = (e^(jwt) + e^(-jwt))*(jwC)/(1 + (RwC)^2)

To put it in exponential form, you can use Euler's formula e^(jwt) = coswt + jsinwt. This will give you:

Is = (coswt + jsinwt + cos(-wt) + jsin(-wt))*(jwC)/(1 + (RwC)^2)

Simplifying this further, you can get:

Is = (2*coswt)*(jwC)/(1 + (RwC)^2)

This is the simplest form for Is.