Question

A 1.50-\(\mu\)F capacitor is charging through a 12.0-\(\Omega\) resistor using a 10.0-V battery. What will be the current when the capacitor has acquired \\(\frac{1}{4}\\) of its maximum charge? Will it be \\(\frac{1}{4}\\) of the maximum current?

Short answer

The current is 0.625 A; it is not \(\frac{1}{4}\) of the maximum current.

Step-by-step solution

Determine Maximum Charge

The maximum charge \(Q_{max}\) that the capacitor can acquire is determined by the equation \(Q_{max} = C \times V\), where \(C = 1.50 \times 10^{-6} \text{ F}\) and \(V = 10.0 \text{ V}\). Calculate \(Q_{max}\): \[ Q_{max} = (1.50 \times 10^{-6} \text{ F}) \times (10.0 \text{ V}) = 15.0 \times 10^{-6} \text{ C} = 15.0 \mu\text{C}. \]

Calculate Quarter Charge

The charge when the capacitor has acquired \(\frac{1}{4}\) of its maximum charge is equal to \(\frac{1}{4} \times Q_{max}\). \[ Q = \frac{1}{4} \times 15.0 \mu\text{C} = 3.75 \mu\text{C}. \]

Calculate Current at Quarter Charge

The current \(I\) in the circuit when the charge \(Q\) on the capacitor is \(3.75 \mu\text{C}\) can be found using the formula for charging a capacitor: \[ I = I_0 e^{-t/RC}, \]where \(I_0 = \frac{V}{R} = \frac{10.0 \text{ V}}{12.0 \Omega} \approx 0.833 \text{ A}.\) Calculate the voltage across the capacitor \(V_c = \frac{Q}{C}\) and the time constant \(RC\), then use these to solve for \(I\).1. Voltage across the capacitor: \[ V_c = \frac{3.75 \times 10^{-6} \text{ C}}{1.50 \times 10^{-6} \text{ F}} = 2.5 \text{ V}. \] 2. Use Kirchhoff's loop rule: \( V = V_c + (I \cdot R) \), rearrange for \(I\): \[ I = \frac{V - V_c}{R} = \frac{10.0 \text{ V} - 2.5 \text{ V}}{12.0 \Omega} \approx 0.625 \text{ A}. \]

Check If Current is a Quarter of Maximum Current

The maximum current \(I_0\) was calculated as approximately \(0.833 \text{ A}\). Determine if the current \(0.625 \text{ A}\) is equal to \(\frac{1}{4}\) of \(I_0\):\[ \frac{1}{4} \times 0.833 \text{ A} = 0.208 \text{ A}. \]The current at \(\frac{1}{4}\) charge \(0.625 \text{ A}\) is not \(\frac{1}{4}\) of the maximum current \(0.208 \text{ A}\).

Key concepts

Capacitor Charging

When a capacitor is connected to a power source, such as a battery, it begins to store electrical energy in an electric field between its plates. This process is known as capacitor charging. The maximum charge a capacitor can store, denoted as \(Q_{max}\), is determined by its capacitance \(C\) and the voltage \(V\) applied across it. The relationship is expressed by the formula \(Q_{max} = C \times V\). In the given exercise, with a 1.50-\(\mu\)F capacitor and a 10.0-V battery, the maximum charge is calculated to be 15.0 \(\mu\)C.

As the capacitor charges, the voltage across it increases, reducing the potential difference between the capacitor and the battery, which in turn reduces the charging current. Understanding this behavior is crucial to analyzing circuits that involve capacitors.

Kirchhoff's Loop Rule

Kirchhoff's Loop Rule is pivotal in understanding electric circuits. This rule states that the sum of the electromotive forces (emfs) and potential differences (voltages) around any closed loop in a circuit must be zero. Essentially, it ensures energy conservation in electrical circuits.

In the context of a charging capacitor, Kirchhoff's Loop Rule can be applied to calculate the current at any given time. For instance, if the voltage across the capacitor at a certain charge is known, the voltage drop across a series resistor can be found using the loop rule: \( V = V_c + I \cdot R \). Rearranging this formula allows for solving the current \(I\), providing insight into the dynamics of the charging process.

This loop rule thus becomes a fundamental tool in circuit analysis, guiding us to understand voltage and current distribution across the circuit elements.

Time Constant RC

The time constant \(RC\) of an RC circuit is a crucial factor in determining how quickly a capacitor charges or discharges. It is the product of the resistance \(R\) and capacitance \(C\): \( \tau = R \times C \). This constant gives us an idea of the time it takes for the charge, voltage, or current to change significantly.

In practical terms, the time constant tells us that after a time equal to \(\tau\), the charging capacitor will have reached about 63% of its maximum charge. Similarly, it describes the time taken for the charge to fall to about 37% of its original value if the capacitor is discharging. For the given exercise, with \(R = 12.0 \Omega\) and \(C = 1.50 \times 10^{-6}\) F, calculate \( \tau \) as \( 18.0 \times 10^{-6} \) seconds.

Understanding the time constant helps us model how quickly or slowly the electrical parameters reach their intended values, crucial for designing circuits with specific timing requirements.

Current and Voltage Calculations

Calculating current and voltage in an RC circuit allows us to predict circuit behavior accurately. At any given time, the current \(I\) through a charging capacitor can be determined using the formula \(I = I_0 e^{-t/RC}\), where \(I_0\) is the initial current, \(R\) is resistance, and \(C\) is the capacitance.

In the exercise, we start by finding the initial current \(I_0 = \frac{V}{R}\), where \(V\) is the battery voltage. For a 10.0-V battery and a 12.0-\(\Omega\) resistor, \(I_0\) is approximately 0.833 A. Then, voltage across the capacitor \(V_c\) is \( \frac{Q}{C} \), with \(Q\) as the charge at a given time. Given \(Q = 3.75 \) \(\mu\)C, \(V_c = 2.5\) V. Using Kirchhoff's Rule, the current is calculated as \(I = \frac{V - V_c}{R}\), which results in 0.625 A.

These calculations are vital for predicting how the current and voltage will vary over time, allowing us to understand and design efficient circuits.