Question

An emf source with \(\varepsilon =\) 120 V, a resistor with \(R =\) 80.0 \(\Omega\), and a capacitor with \(C =\) 4.00 \(\mu\)F are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

Short answer

The charge on each plate of the capacitor is \(1.92 \times 10^{-4} \, C\).

Step-by-step solution

Understanding Circuit Components

The circuit contains an emf source, a resistor, and a capacitor connected in series. The problem gives us the values: \( \varepsilon = 120 \, V \), \( R = 80.0 \, \Omega \), \( C = 4.00 \times 10^{-6} \, F \), and the current \( I = 0.900 \, A \). To find the charge on the capacitor, we use the relationship between voltage, current, and charge.

Determine Voltage Across Resistor

Using Ohm's Law, \( V = IR \), we calculate the voltage drop across the resistor. Substituting the given values, \( V = 0.900 \, A \times 80.0 \, \Omega = 72.0 \, V \).

Calculate Voltage Across Capacitor

The total emf is divided between the resistor and the capacitor. The voltage across the capacitor is \( V_C = \varepsilon - V \), where \( V \) is the voltage across the resistor. Thus, \( V_C = 120 \, V - 72.0 \, V = 48.0 \, V \).

Calculate Charge on Capacitor

Using the formula \( Q = CV \), where \( C \) is the capacitance and \( V \) is the voltage across the capacitor, we find the charge. Substituting the given values, \( Q = (4.00 \times 10^{-6} \, F)(48.0 \, V) = 1.92 \times 10^{-4} \, C \).

Key concepts

Series Circuit

In a series circuit, electrical components are connected end-to-end, which means there is only one path for electric current to flow. This is important because in a series circuit, the same current passes through each component sequentially. Therefore, if one component fails, the circuit will be broken, stopping the current from flowing altogether. Components in a series split the total voltage among them. This division occurs proportionately to their resistances or reactances. Let's consider a circuit with an electromotive force (emf) source, a resistor, and a capacitor. All three are placed one after another such that the current first passes through the emf source, then through the resistor, and finally through the capacitor. Another important point to understand about a series circuit is how the total resistance is calculated. It is simply the sum of individual resistances:

• Total Resistance: If you have multiple resistors connected in series, the total resistance (R_total) is computed as the sum of each resistor: \[ R_{total} = R_1 + R_2 + R_3 + \ \text{... (and so on)} \]

The same current still moves through these resistors despite having varied resistance values, since there is only one path of flow in a series circuit.

Ohm's Law

Ohm's Law is the fundamental principle used to relate voltage, current, and resistance in electrical circuits, and it is given by the equation \( V = IR \), where \( V \) is the voltage across a component, \( I \) is the current flowing through it, and \( R \) is its resistance. This law is quite valuable when analyzing circuits, enabling electricians and engineers to determine any one of the three variables if the other two are known. For example, in our exercise, we used Ohm's Law to find the voltage across a single resistor in a series circuit. Knowing the current (0.900 A) and the resistance (80 \( \Omega \)), the drop in voltage across the resistor was computed as:

• Voltage across Resistor: \[ V = I \cdot R = 0.900 \, A \times 80.0 \, \Omega = 72.0 \, V \]

This concept can be expanded to any number of components in series. The calculated voltage allows us to determine how the other components, like capacitors, react to the circuit’s total emf.

Capacitance

Capacitance is an electrical property that indicates a capacitor's ability to store energy in an electric field and is defined by the equation \( Q = CV \), where \( Q \) is the charge stored (in coulombs), \( C \) is the capacitance (in farads), and \( V \) is the voltage across the capacitor. Understanding capacitance helps us determine how much charge a capacitor can hold at a given voltage. In the previous example, it was crucial to understand the voltage drop differences to find out the charge stored in the capacitor:

• Charge on Capacitor: The voltage across the capacitor was 48.0 V, and with a capacitance of 4.00 \( \mu \)F, the charge \( Q \) was calculated as: \[ Q = (4.00 \times 10^{-6} \, F)(48.0 \, V) = 1.92 \times 10^{-4} \, C \]

This demonstrates how capacitors in a circuit do not just rely on their properties but also on the characteristics of the circuit they reside in. Importantly, they must often be analyzed in the context of other components, such as resistors, to understand the full behavior in a series circuit.