An AC source operates at a maximum voltage of 120 V and an angular frequency of 377 s -1 . If this source is connected in parallel to a 15 Ω resistor and in parallel to a 0.20 mF capacitor, answer the following questions: a) What is the maximum current produced by the source? b) What is the maximum current through the resistor? c) What is the maximum current through the capacitor?

Hey, guys. I hope you're able to solve this problem on your own. If not, here's a little bit of help. An AC source operates at a maximum 120 volts and an angular frequency of 377 inverse seconds. If the source is connected in parallel to a 15 ohm resistor and in parallel to a 0.2 millifarad capacitor, answer the following questions. A, what is the maximum current produced by the source? B, what is the maximum current through the resistor? And c, what is the maximum current through the capacitor? So a, the maximum current produced by the source is always the maximum voltage produced by the source divided by the impedance. Now as we saw in the previous problem, the impedance for a parallel rc circuit sorry, I I should say 1 over the impedance for a parallel rc circuit was the square root of 1 over r squared plus omega squared c squared. Okay? And this is going to lead us sorry. Let me plug in 15 ohms for the resistor plus 377 squared times 0.2 millifarad or 0.2 times 10 to the negative 3 squared. And this is about 0.1 inverse ohms or the impedance is about 10 ohms. Okay? This means that the maximum current in the circuit is just the maximum voltage produced which is a 120 volts divided by 10 ohms or 12 amps. Okay? Part a, very straightforward. Now, part b and part c. The maximum current produced through any element is always going to be the voltage across that element divided by the reactants of that element. Because all of these elements are in parallel the maximum voltage across all of them is equal to the maximum voltage produced by the source. The reactance of a resistor is just r so this is 120 divided by 15 ohms which is 8 amps. And part c, the maximum current through the capacitor is going to be, once again, the maximum voltage across the capacitor which in this case is the maximum voltage produced by the source. This is because they are in parallel divided by the capacitive reactants. Okay? And the capacitive reactants, don't forget, is equal to 1 over omega c. So this whole thing becomes omega c v max which is 377 times 0.2 millifarads times a 120 volts which is the maximum voltage produced by the source which is 9 amps. Okay so let's look we have 3 results. We have 12 amps which is the maximum current through the circuit, sorry, produced by the source, 8 amps which is the maximum current through the, resistor, and 9 amps which is the maximum current through the capacitor. So our circuit looks like this. We have an ac source connected in parallel to a resistor and connected in parallel to a capacitor, and it's producing a current of 12 amps maximum that splits into a maximum current of 8 amps and a maximum current of 9 amps. Notice that this doesn't follow any of the same rules as DC circuits. That's because the maximum currents do not occur at the same time but the rule that it does follow is that these maximum currents should actually be related to each other by the pythagorean theorem that the square root of 8 amps squared plus the square root of 9 sorry plus 9 amps squared that holding square rooted should equal the maximum current produced by the circuit and it actually does equal 12 amps. Okay? So this is the relationship sort of the junction rule in AC circuits not that the maximum current produced by a source equals the sum of the maximum currents through each of the elements. Alright guys, thanks for watching.

Struggling with Physics with Algebra?

Watch next