# 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 questions in mind:

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

2. Is 80 vrms interpreted as 80 V?