91Ó°ÊÓ

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, $$ Q=C V_{a b}=\left(2 \times 10^{-6} \mathrm{~F}\right)(0.25 \mathrm{~V})=0.5 \mu \mathrm{C} $$

Step by step solution

01

Identify Given Information

We are given that the currents in the circuit are steady, with a known capacitor acting as an open circuit since its charge is constant, implying that the current through it, \( I_2 = 0 \). This simplifies the circuit analysis.

02

Calculate Current \( I_1 \)

The circuit can be simplified to a single resistor of \( 12 \, \Omega \) connected to a \( 15 \, \text{V} \) battery since \( I_2 = 0 \). Thus, we use Ohm's Law to find \( I_1 \):\[ I_{1} = \frac{\varepsilon}{R} = \frac{15 \, \text{V}}{12 \, \Omega} = 1.25 \, \text{A} \]

03

Determine Current \( I_3 \)

With \( I_2 = 0 \), the current running through the remaining path of the circuit is the same as \( I_1 \): \[ I_3 = I_1 = 1.25 \, \text{A} \]

04

Calculate Voltage Across Capacitor

To find the voltage difference across the capacitor plates from point \( a \) to point \( b \), follow the voltage changes:\[ \Delta V = -(5.0 \, \Omega \times 1.25 \, \text{A}) + 6.0 \, \text{V} + (3.0 \, \Omega \times 0) = -0.25 \, \text{V} \]

05

Find Charge on Capacitor

Given the voltage across the capacitor \( V_{ab} = 0.25 \, \text{V} \), use the definition of capacitance to find the charge:\[ Q = C \cdot V_{ab} = \left(2 \times 10^{-6} \, \text{F}\right)(0.25 \, \text{V}) = 0.5 \, \mu \text{C} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law

Ohm's Law is a fundamental principle in circuit analysis that relates voltage, current, and resistance in an electrical circuit. It states that the current flowing through a conductor is directly proportional to the voltage across it, and inversely proportional to its resistance. Mathematically, it is expressed as:

• \( I = \frac{V}{R} \)

where:

• \( I \) is the current in amperes (A),
• \( V \) is the voltage in volts (V),
• \( R \) is the resistance in ohms (\(\Omega\)).

In this exercise, Ohm's Law is applied to find the current \( I_1 \) by using the voltage of the battery (15 V) and the resistance (12 \(\Omega\)). This shows how useful Ohm's Law is for determining current when voltage and resistance are known. By understanding this law, you can analyze a wide range of basic circuits.

Steady Current

A steady current means that the current flow is constant and does not change with time. In circuit analysis, steady-state conditions imply that all currents and voltages are constant over time, having reached equilibrium.
In such circuits:

• All reactive components like inductors and capacitors have no changing stored energy.
• Inductors are replaced by short circuits, while capacitors are replaced by open circuits (or vice versa if changes are non-uniform).

In the provided exercise, \( I_2 = 0 \) because the capacitor is fully charged, resulting in a steady-state condition. This means the current bypasses the path through the capacitor, and the rest of the circuit operates as usual. Understanding steady current conditions helps simplify complex circuits into manageable parts for analysis.

Charge on Capacitor

Capacitors store electrical energy in the form of electrostatic charge. The charge \( Q \) on a capacitor is determined by the relationship:

• \( Q = C \cdot V \)

where:

• \( Q \) is the charge in coulombs (C),
• \( C \) is the capacitance in farads (F),
• \( V \) is the voltage across the capacitor in volts (V).

In this exercise, the charge on the capacitor is calculated using the known capacitance (2 \(\times 10^{-6}\) F) and the voltage difference across it (-0.25 V, but the magnitude is used). This shows that the charge is a function of both the capacitance and the voltage difference across the capacitor. Understanding how to calculate the charge on a capacitor is crucial for analyzing circuits where energy storage is involved.