Question

The graph shows the voltage across and the current through a single circuit element connected to an ac generator. Identify the circuit element The element is a 25-ohm resistor. The element is a 35-ohm resistor. The element is a 0.45-H inductor. The element is a 360-mu F capacitor. The element is a 510-mu F capacitor.

could someone please explain the whole frequency concept here and how they got it?

Answer 1

As in the graph, we can see that current is leading the voltage by 90 degrees. Means currents reach its peak value before voltage in one-fourth of a cycle. As one cycle means 360 degrees. 90 degrees represent one-fourth of a cycle.

We all know that currents lead by 90 degrees for the capacitor. So given graph is definitely for the capacitor.

So, by ohms law: -

As in the graph, we can see that current is leading the voltage by 90 degrees. Means currents reach its peak value before voltage in one-fourth of a cycle. As one cycle means 360 degrees. 90 degrees represent one-fourth of a cycle. We all know that currents lead by 90 degrees for the capacitor. So given graph is definitely for the capacitor. So, by Ohm law: V = IR implies R = V/I So, R = 99/3.8 = 34.4 Ohm As we can see in graph that current completed 1.75 cycles in 0.14 sec means T = 0.14/1.75 = 0.8 Sec Resistance for capacitor is denoted by X_c and given by X_c = 1/2 pi f C So. C = 1/2 pi f X_c and f = 1/T C = 1/2 pi times 12.5 times 35.4 = 3.597 times 10^-4 F C almostequalto mu F

So. $R = \frac{99}{3.8} = 34.4 ohms$

As we can see in graph that current completed 1.75 cycles in 0.14sec means

Resistance for capacitor is denoted by $X_{c}$ and given by $X_{c} = \frac{1}{2\pi f C}$

So. $C = \frac{1}{2\pi f X_c}$ and $f = \frac{1}{T}$

$C = \frac{1}{2\pi *12.5 *35.4} = 3.597*10^{-4}F$

$C \approx 360\mu F$