- #1

zenterix

- 563

- 77

- Homework Statement
- An AC source is connected to a single unknown circuit element as shown in the second picture below.

The driving frequency is ##\omega=100\text{rad/s}##.

The voltage and current are as in the plot in the third picture below.

- Relevant Equations
- Which of the following statements could be true

Here is the circuit

and here is the plot of current and voltage

we don't know which is which initially.

Just by looking at this plot, I conclude that the element cannot be a resistor because if it were then the phase would need to be zero.

Next, suppose the element is an inductor. Then

$$I_{L0}=\frac{V_{L0}}{\omega L}$$

where ##I_{L0}## and ##V_{L0}## are amplitudes of current and voltage for such a circuit.

Now, but visual inspection of the plot we see that we can have two cases.

Suppose the dashed graph is the current and the solid graph is the voltage. Then

$$\mathrm{200mA=\frac{10V}{100rad/s\cdot L}}$$

$$\implies L=\frac{1}{2}\text{H}$$

Next suppose that the dashed graph is the voltage and the solid graph is the current. Then

$$\mathrm{100mA=\frac{20V}{100rad/s\cdot L}}$$

$$\implies L=2\text{H}$$

Next, suppose the element is a capacitor. By analogous reasoning, but now using the equation

$$I_{C0}=\omega C V_{CO}$$

we reach two cases.

If the dashed line is current then we find that

$$\mathrm{200mA=100rad/s \cdot C\cdot 10V}$$

$$\implies C=\frac{0.2}{1000}=0.2\text{mF}$$

If the dashed line is voltage then

$$\mathrm{100mA=100rad/s\cdot C\cdot 20V}$$

$$\implies C=\frac{0.1}{2000}\text{F}=50\mathrm{\mu F}$$

If this is all correct I have shown that the four options selected in the first picture above are correct and that the two unselected options are incorrect. If this is so, the grading system is incorrect.

On the other hand, I guess it is more probable that I am making some mistake.