# Finding the total impedance and voltage across open terminals

The question that I'm trying to solve is the following:

In the circuit shown below, what is the total impedance in ohms across the open terminals? What is $$\v_{T}\$$ if $$\v_{L}\$$ is 80 vrms?

This is how I solved the problem.

I let $$\Z_{1}\$$ be equal to the 10-ohm resistor, $$\Z_{2}\$$ to the -j60-ohm, and $$\Z_{3}\$$ to $$\30+j40\$$.

Solving for the total impedance: $$z_{t} = z_{1}+z_{2}\left | \right |z_{3} = 10 + \frac{\left ( -j60 \right )\left ( 30+j40 \right )}{30-j20}=\frac{1210}{13}-\frac{60}{13}j$$

I'm not sure though if looking for the total impedance is similar to looking for the equivalent impedance/input impedance as shown in my textbook.

For the voltage, I used the voltage division principle:

$$v_{L} = \frac{z_{3}}{z_{1}+z_{2}+z_{3}}\left ( v_{T} \right )$$

Then, I solved for $$\v_{T}\$$ which is equal to $$\\frac{64}{5}-j\frac{352}{5}\$$.

Now, I have two worries in mind.

1. Is total impedance the same as input impedance or equivalent impedance?

2. Is 80 vrms interpreted as 80 V?

Your total impedance is correct.

However your voltage divider is wrong. The correct voltage divider is VL = (j40)/(Z3)*Vx where Vx is the voltage accross Z2 and Z3.

To find this voltage, you can use another voltage divider. Vx = (Z2//Z3)/(Z1 + Z2//Z3) * Vt.