Find the impedance of capacitor and inductor with dependant current so

In summary, the homework statement is trying to find the impedances of the capacitor and inductor. The Attempt at a Solution states that the impedance of the capacitor is -j10 and the impedance of the inductor is j.5.
  • #1
pokie_panda
37
0

Homework Statement



From the circuit find the impedances of the capacitor and inductor

attachment.php?attachmentid=58273&stc=1&d=1367084378.jpg


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


Homework Equations





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
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
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.
 
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  • #4
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
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
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
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
sorry i should of said delta , its the current across the capacitor.
So this should have a angular frequency of 500
 
  • #9
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
pokie_panda said:
sorry i should of said delta , its the current across the capacitor.
So this should have a angular frequency of 500

That's right.
 
  • #11
Thanks for the help, sorry for posting in the wrong area.
 

1. What is impedance?

Impedance is a measure of the opposition to current flow in an electrical circuit. It is represented by the symbol Z and is measured in ohms (Ω).

2. How do you calculate the impedance of a capacitor?

The impedance of a capacitor can be calculated using the formula Z = 1/(jωC), where j is the imaginary unit, ω is the angular frequency, and C is the capacitance in farads.

3. What is the relationship between impedance and frequency for a capacitor?

The impedance of a capacitor decreases as the frequency of the current passing through it increases. This is because the capacitor acts as a short circuit for high frequency signals.

4. How do you find the impedance of an inductor?

The impedance of an inductor can be calculated using the formula Z = jωL, where j is the imaginary unit, ω is the angular frequency, and L is the inductance in henries.

5. What is the effect of frequency on the impedance of an inductor?

The impedance of an inductor increases as the frequency of the current passing through it increases. This is because the inductor acts as a short circuit for low frequency signals and an open circuit for high frequency signals.

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