Impedance of circuit with capacitor

[athenaroa]

You are given a capacitor( impedance 100 ohms), an inductor (impedance 300 ohms) and a resistor (400 ohms), all connected in series. Determine the impedance( in ohms) of this circuit.

 

[Averagesupernova]

This sounds as if you are trying to get someone to do your homework. Can you give us a little more detail? Before we give you an answer, can you at least give us several possible outcomes of the problem?

 

[Chen]

(Assuming your EMF is of AC:)

$${V_r}_{max} = {I_s}_{max}r$$
$${V_L}_{max} = {I_s}_{max}X_L$$
$${V_c}_{max} = {I_s}_{max}X_c$$

I'm not going to explain the whole theory here, but you should know that in this kind of circuit the $$V_c$$ precedes the current by 90 degrees, $$V_L$$ is lagging by 90 degrees, and $$V_r$$ corresponds to the current exactly. (I'm probably using the wrong terms and not explaining this very well ).

Therefore you can draw a diagram of $${V_c}_{max}$$, $${V_L}_{max}$$, and $${V_r}_{max}$$, whereby $${V_c}_{max}$$ is 90 degrees ahead of $${V_r}_{max}$$ and $${V_L}_{max}$$ is 90 degrees behind it.

But you should also know that all the time:

$$V_r + V_L + V_c = V_s$$

So to find $$V_s$$ at any moment you need to sum up all three V's, while treating them as vectors. On the axis that connects $${V_c}_{max}$$ and $${V_L}_{max}$$ the sum would be $${V_c}_{max} - {V_L}_{max}$$, and on the other axis the sum would simply be $${V_r}_{max}$$. Using pythagoras you can show that:

$$V_s = \sqrt{({V_c}_{max} - {V_L}_{max})^2 + {V_r}_{max}^2}$$

Which becomes:

$${I_s}_{max}Z = \sqrt{({I_s}_{max}X_c - {I_s}_{max}X_L)^2 + ({I_s}_{max}r)^2}$$

And if you divide by $${I_s}_{max}$$:

$$Z = \sqrt{(X_c - X_L)^2 + r^2}$$

So Z, or the impedance of this circuit, is $$\sqrt{200k\Omega}$$