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?

enter image description here

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?


1 Answer 1


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.

To answer your two questions:

  1. The "total impedance" has no real meaning. Rather we will speak about a total impedance accross some terminals. The input impedance is just the total impedance accross the input terminals. It depends on where you consider your input terminals are.

  2. It depends on what you mean as "80V". 80 VRMS is still a voltage, but is different from for example the amplitude of your signal. Assuming you are speaking about a sine wave, you have the following formula: A = sqrt(2)*VRMS where A is the amplitude of your sine wave.

  • \$\begingroup\$ By saying that V_x is multiplied in your voltage divider in VL means that V_x should be mulitplied as a numerator, not as a denominator, right? Hehehe. \$\endgroup\$
    – romeoPH
    Commented Jun 19, 2020 at 8:09
  • \$\begingroup\$ Well, by definition of a voltage divider, I multiplied vx and vt as a numerator. I found the vt already: (90-j(200/3))V. \$\endgroup\$
    – romeoPH
    Commented Jun 19, 2020 at 8:15

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