Question

How long will it take for the current in a circuit to drop from its initial value to \(1.50 \mathrm{~mA}\) if the circuit contains two \(3.8-\mu \mathrm{F}\) capacitors that are initially uncharged, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) battery all connected in series?

Short answer

Answer: It will take approximately 4.58 ms for the current in the circuit to drop to 1.50 mA.

Step-by-step solution

Determine the equivalent resistance and capacitance

Since the two capacitors and the two resistors are connected in series, we will calculate the equivalent resistance (R) and capacitance (C) of the circuit. For resistors connected in series, we can simply add their resistance values together: \(R_{eq} = R_1 + R_2 = 2.2\,\text{k}\Omega + 2.2\,\text{k}\Omega = 4.4\,\text{k}\Omega\) For capacitors connected in series, the inverse of the equivalent capacitance is equal to the sum of the inverses of the two capacitances: \(\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}\) Insert the values of the given capacitances: \(\frac{1}{C_{eq}} = \frac{1}{3.8\,\mu\text{F}}+\frac{1}{3.8\,\mu\text{F}} = \frac{2}{3.8\,\mu\text{F}}\) Now, we can find the equivalent capacitance (C) as follows: \(C_{eq} = \frac{1}{\frac{2}{3.8\,\mu\text{F}}} = 1.9\,\mu\text{F}\)

Calculate the time constant

The time constant (τ) of an RC circuit can be calculated using the formula: \(\tau = RC\) Insert the values of the equivalent resistance and capacitance: \(\tau = (4.4\,\text{k}\Omega)(1.9\,\mu\text{F}) = 8.36\,\text{ms}\)

Use the formula for the charging current

The charging current (I) in the circuit after a time t can be given by the formula: \(I(t) = I_0 e^{-t/\tau}\) Where \(I_0\) is the initial current, and t is the time. The initial current in the circuit can be found using Ohm's law: \(I_0 = \frac{V}{R_{eq}}=\frac{12.0\,\text{V}}{4.4\,\text{k}\Omega}=2.73\,\text{mA}\) We will now insert the given final current value, \(1.50\,\text{mA} \), and solve for t: \(1.50\,\text{mA} = 2.73\,\text{mA} e^{-t/8.36\,\text{ms}}\)

Solve for the required time t

First, divide both sides by \(2.73\,\text{mA}\): \(\frac{1.50\,\text{mA}}{2.73\,\text{mA}} = e^{-t/8.36\,\text{ms}}\) Now, take the natural logarithm of both sides: \(-t/8.36\,\text{ms} = \ln\left(\frac{1.50\,\text{mA}}{2.73\,\text{mA}}\right)\) Finally, solve for t: \(t = -8.36\,\text{ms} \times \ln\left(\frac{1.50\,\text{mA}}{2.73\,\text{mA}}\right) \approx 4.58\,\text{ms}\) Therefore, it will take approximately 4.58 ms for the current in the circuit to drop from its initial value to 1.50 mA.

Key concepts

Time Constant

In the study of RC circuits, understanding the concept of a time constant is essential. The time constant, often denoted by the Greek letter \( \tau \), is a measure of how quickly an RC circuit responds to changes in voltage or current. It is calculated as the product of the equivalent resistance \( R \) and the equivalent capacitance \( C \) of the circuit. Therefore, \( \tau = RC \).

The time constant is expressed in units of time, such as seconds or milliseconds, and indicates how long it takes for a capacitor to charge or discharge to approximately 63.2% of its full charge. Conversely, it is also the time it takes for the current to decrease to about 36.8% of its original value when discharging.

The higher the time constant, the longer it takes for the capacitor to charge or discharge. In practical terms, five time constants are usually required for a capacitor to be charged or discharged almost completely (more than 99%). Understanding the time constant is therefore critical in designing circuits that need precise timing.

Ohm's Law

Ohm's Law is a foundational principle in electrical engineering that describes the relationship between voltage, current, and resistance in a circuit. It is mathematically expressed as \( V = IR \), where \( V \) is the voltage across a component, \( I \) is the current flowing through it, and \( R \) is its resistance.

In the context of an RC circuit, Ohm’s Law can be applied to determine the initial current flowing in the circuit when the capacitors are uncharged. By knowing the total voltage supplied by the battery and the equivalent resistance of the resistors in series, the initial current can be easily calculated. For instance, if you have a \( 12.0 \mathrm{~V} \) battery and the equivalent resistance is \( 4.4\,\text{k}\Omega \), then the initial current \( I_0 \) is given by \( I_0 = \frac{12.0\,\text{V}}{4.4\,\text{k}\Omega} \), resulting in \( 2.73\,\text{mA} \).

This principle is crucial for understanding how voltage and current distribute across different components of a circuit and is particularly useful for troubleshooting and designing electrical circuits.

Exponential Decay

Exponential decay is a mathematical concept that describes the process by which a quantity decreases at a rate proportional to its current value. In RC circuits, the current and voltage across the capacitor follow an exponential decay over time as the circuit responds to discharging events.

This phenomenon is described by the formula for the current in a discharging RC circuit: \( I(t) = I_0 e^{-t/\tau} \), where \( I_0 \) is the initial current and \( \tau \) is the time constant. This formula shows how the current decreases exponentially from its initial value over time.

For example, if an RC circuit starts with an initial current of \( 2.73\,\text{mA} \) and is supposed to drop to \( 1.50\,\text{mA} \), this exponential model allows us to calculate the time required for such a reduction by finding the time \( t \) using the natural logarithm of the ratio of final to initial current values. This characteristic of exponential decay is fundamental to understanding the dynamics of electric circuits involving capacitors.

Equivalent Resistance and Capacitance

In series circuits that include resistors and capacitors, finding the equivalent resistance and capacitance is a necessary step to analyze the circuit's behavior effectively.

When resistors are connected in series, as in the given problem, the total or equivalent resistance \( R_{eq} \) is simply the sum of the individual resistances: \( R_{eq} = R_1 + R_2 \). So for two \( 2.2\,\text{k}\Omega \) resistors, \( R_{eq} = 4.4\,\text{k}\Omega \).

Capacitors behave oppositely when connected in series. The inverse of the equivalent capacitance \( C_{eq} \) is equal to the sum of the inverses of each capacitor's capacitance. For capacitors \( C_1 \) and \( C_2 \), connected in series:

• \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \)

Thus, for two \( 3.8\,\mu\text{F} \) capacitors in series, the \( C_{eq} \) would be \( 1.9\,\mu\text{F} \).

Understanding these concepts helps simplify the analysis of complex circuits since it reduces them into simpler equivalent circuits for easier calculations.