The circuit in Figure P 28.37 has been connected for a long time. (a) What is the voltage across

The circuit in Figure P 28.37 has been connected for a long...

Key Concepts

Steady-State Behavior of Capacitors

In DC circuits that have been connected for a long time, the capacitor reaches a steady state where it is fully charged and behaves like an open circuit. This means that the voltage across the capacitor equals the voltage of the connected source (accounting for any voltage drops in the circuit), and no further current flows through the capacitor branch.

RC Time Constant

The RC time constant, denoted by ? (tau), is the product of the resistance (R) and the capacitance (C) in an RC circuit. It characterizes how quickly a capacitor charges or discharges: after a time period equal to ?, the voltage across a discharging capacitor decreases to about 37% of its initial value. This concept is crucial for analyzing transient behavior in RC circuits.

Exponential Decay in Capacitor Discharge

When a capacitor discharges through a resistor, the voltage across it decays exponentially according to the equation V(t) = V? e^(?t/RC), where V? is the initial voltage. This relationship allows us to calculate the time needed for the voltage to drop to a specific fraction of its original value by solving the exponential equation, which is fundamental in transient circuit analysis.

Transcript

00:01 Okay, so i have this configuration and i want to know what is the potential difference across the capacitor.

00:06 And what happens when the battery is disconnected, how long it takes for the capacitor to discharge to 1 -10 of the initial valve? okay, so we will start with the first part.

00:21 So i want you to notice first here, we have a current, okay, a current that flows from negative to positive.

00:29 It really does not matter much the sense.

00:32 And when it comes to this point, it will divide, because this is a parallel configuration, okay? so it will divide between i -1 and i -2.

00:42 Okay, you can see that actually r1 and r2 are in a serious configuration.

00:48 They both have i -1, and r3 and r4 have equally the same series configuration, and the current flow through here, flows, to both of them, i too.

01:02 And actually, since the battery has been connected for a long time, the current across the capacitor is zero.

01:11 It means it's fully charged and my system is an equilibrium.

01:15 So actually, i can just think that this does not exist for now.

01:20 Okay, so what is the potential difference there? okay, the first thing i have to find is the current.

01:29 Okay, this will be useful and you will see why.

01:33 So i know that to find i -1, okay, for omslaw, this will be the potential difference divided by the resistance.

01:45 And i know, remember that for a parallel configuration, the potential difference is the same, and since the 10 volt is the potential difference, this will be 10 volts and the total resistor value is going to be the sum of the 1r1 and r2.

02:01 Okay, i'm just going to put it like this.

02:04 This is because it's a series configuration.

02:07 Okay, so actually this is 10 divided by 5 and this is 2 amp.

02:12 So i found the value of i1 and i do exactly the same for r2.

02:18 So this is 10 divided by r3 plus r4 and this is 10 divided by 10.

02:26 So it is 1 ampers.

02:28 Okay, so how do i find the potential difference? i have to find first from here, let's call this point a.

02:39 And then to here, let's call this point b.

02:41 So i will choose a path to follow from point to point.

02:48 I will actually choose this path.

02:51 Okay, you can choose the path below.

02:55 It will yield the same result.

02:57 Okay, so you will see that the first thing i find is what is the potential difference in r1.

03:06 Okay, i'm going to call this v1.

03:09 Okay, i do not know, but i know that in a series configuration, this is this potential difference is going to divide.

03:20 Okay, and actually since the sum of both of the potentials in r1 and r2, must be 10.

03:29 So i know the potential difference in 1 must be 10 minus and then i apply omenslow.

03:36 Okay, remember i never forget.

03:39 And this is what this will be the value of the resistor, which is 1.

03:46 Multiplying the current and i know that i 1 is 2 ampere...

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