The currents in the circuit are steady. Find I1, I2, I3, and...

The currents in the circuit are steady. Find $I_{1}, I_{2}, I_{3}$, and the charge on the capacitor. When a capacitor has a constant charge, as it does here, the current flowing to it is zero. Therefore, $I_{2}=0$, and the circuit behaves just as though the center wire were missing. With the center wire missing, the remaining circuit is simply 12 $\Omega$ connected across a 15 -V battery. Therefore, $$ I_{1}=\frac{\varepsilon}{R}=\frac{15 \mathrm{~V}}{12 \Omega}=1.25 \mathrm{~A} $$ In addition, because $I_{2}=0$, we have $I_{3}=I_{1}=1.3 \mathrm{~A}$. To find the charge on the capacitor, first find the voltage difference between points- $a$ and $-b$. Start at $a$ and go around the upper path. Voltage change from $a$ to $b=-(5.0 \Omega) I_{3}+6.0 \mathrm{~V}+(3.0 \Omega) I_{2}$ $$ =-(5.0 \Omega)(1.25 \mathrm{~A})+6.0 \mathrm{~V}+(3.0 \Omega)(0)=-0.25 \mathrm{~V} $$ Therefore, $b$ is at the lower potential and the capacitor plate at $b$ is negative. To find the charge on the capacitor, $$ Q=C V_{a b}=\left(2 \times 10^{-6} \mathrm{~F}\right)(0.25 \mathrm{~V})=0.5 \mu \mathrm{C} $$

The currents in the circuit are steady. Find $I_{1}, I_{2}, I_{3}$, and the charge on the capacitor.
When a capacitor has a constant charge, as it does here, the current flowing to it is zero. Therefore, $I_{2}=0$, and the circuit behaves just as though the center wire were missing.
With the center wire missing, the remaining circuit is simply 12 $\Omega$ connected across a 15 -V battery. Therefore,
$$
I_{1}=\frac{\varepsilon}{R}=\frac{15 \mathrm{~V}}{12 \Omega}=1.25 \mathrm{~A}
$$
In addition, because $I_{2}=0$, we have $I_{3}=I_{1}=1.3 \mathrm{~A}$.
To find the charge on the capacitor, first find the voltage difference between points- $a$ and $-b$. Start at $a$ and go around the upper path.
Voltage change from $a$ to $b=-(5.0 \Omega) I_{3}+6.0 \mathrm{~V}+(3.0 \Omega) I_{2}$
$$
=-(5.0 \Omega)(1.25 \mathrm{~A})+6.0 \mathrm{~V}+(3.0 \Omega)(0)=-0.25 \mathrm{~V}
$$
Therefore, $b$ is at the lower potential and the capacitor plate at $b$ is negative. To find the charge on the capacitor,
$$
Q=C V_{a b}=\left(2 \times 10^{-6} \mathrm{~F}\right)(0.25 \mathrm{~V})=0.5 \mu \mathrm{C}
$$

Key Concepts

Calculating Capacitor Charge

The charge stored on a capacitor is determined by the relationship Q = CV, where C is the capacitance and V is the voltage across the capacitor. This fundamental equation allows one to compute the amount of charge accumulated on the capacitor plates once the voltage difference has been determined using circuit analysis techniques.

Ohm's Law

Ohm's Law is a core principle in electrical circuits, stating that the current through a resistor is directly proportional to the voltage across it and inversely proportional to its resistance. It is mathematically expressed as I = V/R, and it is used to determine the current in resistive elements within a circuit.

Kirchhoff's Loop Law

Kirchhoff's Voltage Law (KVL) states that the algebraic sum of all voltage changes around any closed loop in a circuit must equal zero. This principle is employed to analyze the voltage drops across various circuit components and to relate them to the overall applied voltage.

Steady-State Analysis in Circuits

In a steady-state circuit, all the currents and voltages have become constant over time. This means that transient behaviors like charging or discharging processes have finished, and any capacitors in the circuit reach a final state where they effectively behave as open circuits for DC analysis.

Capacitor Behavior in DC Circuits

When a capacitor is fully charged in a DC circuit, no current flows through it because it acts as an open circuit. This is a fundamental concept, as it simplifies circuit analysis by allowing one to ignore any branch that solely contains a capacitor under steady conditions.

Transcript

00:07 In this lesson, we're looking at a steady state rc circuit and we want to find out a couple of things.

00:12 We want to find the currents at a few points and we want to find the charge on a capacitor.

00:17 So let me draw you the circuit.

00:20 Okay.

00:22 So you can see here we have a simple circuit.

00:24 We have a 15 volt battery, right, which is providing a current.

00:29 We have three resistors.

00:31 We have five ooms here, seven ooms down here, three oms here.

00:36 Right here we have a two microferriceric capacitor.

00:41 We have another six -volt battery.

00:44 And what we want to find is the currents at point i -1, i -2 here, and i -3.

00:52 And we want to find the charge on our capacitor, on our two micro -faird capacitors.

00:57 All right, so we're given that the current is in the steady state, which means the charge on the capacitor is constant.

01:03 What can we tell from that? we know that the current going through the two capacitor will be zero.

01:10 So we can cut out this middle part here since there's no current flowing through it.

01:14 We have i2 equals zero.

01:17 All right.

01:19 And then what we have is our circuit can be simplified to a simple circuit consisting of a voltage supply or battery and two resistors in series.

01:31 Right.

01:32 And now we see that i1 equals i3.

01:34 We still have i2 equals zero.

01:39 All right.

01:39 How do we find these two currents? we use our equivalent resistance formula to find the total resistance of the circuit.

01:47 They're in series.

01:48 So our resistance is just the sum.

01:50 Our eq equals five ooms plus seven ooms equals 12 oms.

02:00 Right.

02:00 And we know from oms law v equals ir or i equals v over r, which is our 15 volts divided by 12 oms.

02:14 So our current i, that's i equals i1 equals i3, equals 15 volts over 12 oms equals 1 .25 ampers...