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.
Is total impedance the same as input impedance or equivalent impedance?
Is 80 vrms interpreted as 80 V?
Thanks in advanced!