# Find the impedance of capacitor and inductor with dependant current so

1. Apr 27, 2013

### pokie_panda

1. The problem statement, all variables and given/known data

From the circuit find the impedances of the capacitor and inductor

Is(t)=15 cos(500t) A

2. Relevant equations

3. The attempt at a solution

So is this right? Convert the 15 cos 500t into polar form, 15<0 I. Then the resistor is 8 ohms, so 15*8 = 120 V.

Therefore, to find the impedance capacitor is

v(t)=120 cos(500t)
Zc= -1/(0.2*10^-3 * 500)
=-j10

but now I'm stuck. How do find grad is(t)?

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2. Apr 27, 2013

### vela

Staff Emeritus
No, that's not right. I suggest you start by telling us what the formulas are for the impedance of a capacitor and an inductor are, which you should have included in the relevant equations.

3. Apr 27, 2013

### pokie_panda

The relevant equations

Capacitor Impedance= j(-1/wC) using v(t)
Inductor impedance = (jwL) using i(t)

This equations for capacitor you need voltage could even using the i(t)= 15 cos 500t you still get
Zc as= -j10.

Last edited: Apr 27, 2013
4. Apr 27, 2013

### vela

Staff Emeritus
What do you mean "using v(t)" and "using i(t)"? Neither v nor i appear in the formulas. What does $\omega$ stand for?

5. Apr 27, 2013

### pokie_panda

v(t), i(t) are the functions and w is the angular frequency

for example

C= 2uf
v(t)=200cos(5t) mV

Zc= j *-1/((2*10^-6 )(5))

6. Apr 27, 2013

### pokie_panda

Is this the right assumption to make we know that the current is going to be the same,
So
Zc= j -1/(500*(0.2*10^-3))
=-j10
2.5*15 cos(500t) the frequency should be the same ?
ZL=j*500*(10*10-3)
=j.5

7. Apr 27, 2013

### vela

Staff Emeritus
Sorry, brain fart on my part. Your answer for Zc is correct.

Remember that the impedance is the ratio of voltage and current: Z = V/I. It doesn't depend on either V or I; it depends on both. The voltage across an element and the current flowing through it will always have the same frequency. The impedance simply changes the amplitude and phase of the sine wave.

So what did you mean by "grad is(t)"?

8. Apr 27, 2013

### pokie_panda

sorry i should of said delta , its the current across the capacitor.
So this should have a angular frequency of 500

9. Apr 27, 2013

### vela

Staff Emeritus
I should clarify a bit. The way you determined the voltage across the capacitor wasn't correct. The current from the current source doesn't all go through the resistor, so you can't find the voltage across the resistor simply by multiplying is(t) by R.

In this problem it doesn't matter because you don't need the voltage across the capacitor. All you need to know is the angular frequency, which is constant throughout the circuit.

10. Apr 27, 2013

### vela

Staff Emeritus
That's right.

11. Apr 27, 2013

### pokie_panda

Thanks for the help, sorry for posting in the wrong area.