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}