The currents in the circuit in Fig. 28-11 are steady. Find...

The currents in the circuit in Fig. $28-11$ 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, $$ I_{1}=\frac{\mathscr{E}}{R}=\frac{15 \mathrm{~V}}{12 \Omega}=1.25 \mathrm{~A} $$ 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 in Fig. $28-11$ 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,
$$
I_{1}=\frac{\mathscr{E}}{R}=\frac{15 \mathrm{~V}}{12 \Omega}=1.25 \mathrm{~A}
$$
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

Capacitor Charge-Voltage Relationship (Q = CV)

The direct relationship between the charge Q stored in a capacitor, its capacitance C, and the voltage V across it is given by the formula Q = CV. This equation is central in determining how much charge is stored in a capacitor when the voltage across it is established, especially in steady state conditions.

Ohm’s Law

Ohm’s Law defines the relationship between voltage, current, and resistance in an electrical circuit, expressed as V = IR. This fundamental principle is used to calculate the current flowing through resistive elements when the voltage across them is known.

Steady State Analysis in Circuits

In circuit analysis, 'steady state' refers to the condition where all circuit variables (currents, voltages) have stopped changing with time. This is particularly important for circuits involving capacitors or inductors, as their transient behaviors have subsided, leading to simplified analysis based on constant values.

Capacitor Behavior in DC Steady State

A key concept is that a capacitor, when fully charged in a DC circuit, reaches a constant voltage and no current flows through it, effectively acting as an open circuit. This property simplifies circuit analysis since branches that include capacitors can be removed from the current path once steady state is achieved.

Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law states that the sum of all electrical potential differences around any closed loop in a circuit is zero. This principle is crucial in analyzing circuits by forming equations that account for the drops across resistors, voltage sources, and other components in a closed loop.

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