(II) Consider the circuit shown in Fig. 26–67, where all resis...

Question

(II) Consider the circuit shown in Fig. 26–67, where all resistors have the same resistance R. At t = 0, with the capacitor C uncharged, the switch is closed. At t = ∞, what is the potential difference across the capacitor?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with circuits. This problem says in a live experiment, a circuit is constructed with three identical resistors, each with a resistance of 100 ohms and an initially uncharged capacitor at T equal to zero. The switch is closed allowing the current to flow through the circuit. As the system reaches T equal to infinity, the capacitor becomes fully charged and no current flows through the branch containing the capacitor. Your task is to help the researcher determine the potential difference across the capacitor at T equal to infinity. After the system has reached a steady state. Below the question, we're given a circuit diagram for our problem. We're also given four possible choices as our answers. For choice. A, the potential difference across the capacitor at T equal to infinity is E divided by two. For choice B, the potential difference across the capacitor at T equal to infinity is three E divided by two. For choice C, the potential difference across the capacitor at T equal to infinity is three E divided by 200. And for choice D, the potential difference across the capacitor at T equal to infinity is E divided by 200. Now, since we're looking at the state in which T is equal to infinity, and our capacitor is fully charged and we have no current flowing through the branch of the circuit that contains the capacitor. So if I look at my circuit diagram, that means that the current labeled I three, which is flowing through the right branch of our circuit that contains the capacitor, its value is going to be equal to zero amps. That also means that the other two currents in our circuit I one and I two have to be equal. So we have I one equal to I two. That's because I one would be going into our junction. And so whatever is going into the junction needs to come out of the junction. So typically, that'd be I two plus I three. But since I three is zero, that means I one is going to be equal to I two from um Kirk off's junction rule. Now we're tasked to tackling the voltage across the capacitor and the voltage across the capacitor is actually going to be equal to the voltage across our resistor. That's in the middle of our circuit diagram that has I two flowing through it. And that's because the capacitor is in parallel with that resistor. And so if we can figure out what I two is, we can use ohms law to calculate the voltage across that resistor. And then set that voltage equal to the voltage across the capacitor to figure out the current value for I two. We're going to use Kerkhoff S loop rule on the left branch of our circuit. So we're gonna have that loop going around in a clockwise direction. And so we're gonna start at a point below the battery. And so the first element that I wanna come to is going to be our battery. So we'll have a voltage gain of E, the next element that I wanna come to is a R resistor that has I one going through it. And since I'm going in the same direction as the current, that means I'm gonna have a voltage drop. So I'll have minus I one multiplied by the resistance of 100 ohms. I'll continue around the circuit and the next resistor, uh next element that I come to is the resistor that has I two going through it. And I'm going to uh cross this resistor in the same direction as the current. Again, it'll be able to drop. So I'll have minus I two multiplied by 100 ohms. And if I continue around the loop, I'm back to my starting point. So this all has to be equal to zero. So I can make the substitution that I, one is equal to I two and then combine our two terms with the current. So we'll have E minus I two multiplied by 200 ohms. That's gotta be equal to zero. So I can now solve four I two by moving the I two term to the other side and divided by uh 200 ohms. That means that I two is equal to E divided by 200 ohms. So now we can use that current to calculate the voltage across the capacitor because the voltage across the capacitor which will label as VC has to be equal to the voltage across the resistor that has I two going through it. We'll label that as V two. And so we can use OHMS all to calculate V two will have VC is equal to or OHMS. All will have the current I two multiplied by the resistance, which is 100 ohms. We can plug in our expression for I two. And so we'll have VC is equal to E divided by 200 ohms. And that fraction is multiplied by 100 ohms. And so we can simplify that and we'll get VC is equal to E divided by two. So that's the answer I'm looking for. And this sex corresponds to answer choice. A so just a quick little recap of what we've done here. We use Kirkoff loop rule in order to find the current that is going across the res going through the resistor. Um That is in the center of our cir circuit diagram, we then use Ohm's law to calculate the voltage across that resistor using the current I two that we just found and that voltage across that resistor was equal to the voltage across the capacitor because the uh capacitor and resistor are in parallel with each other. So I hope that this has been useful and I'll see you in the next video.

(II) Consider the circuit shown in Fig. 26–67, where all resistors have the same resistance R. At t = 0, with the capacitor C uncharged, the switch is closed. At t = ∞, what is the potential difference across the capacitor?

Key Concepts

Capacitance

Steady State in RC Circuits

Voltage Division in Series Circuits

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