Problem 45 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Two 3.8- \(\mu\) F capacitors, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) source are connected in series. Starting from the uncharged state, how long does it take for the current to drop from its initial value to \(1.50 \mathrm{~mA} ?\)

Step-by-Step Solution

Answer

It takes approximately 5.12 ms for the current to drop to 1.50 mA.

Step 1: Understanding Series RC Circuit

In a series circuit with capacitors and resistors, the time it takes for current to drop to a certain value can be calculated using the RC time constant. The given circuit consists of two identical capacitors and two identical resistors all in series, with an initial voltage source applied.

Step 2: Calculate Equivalent Capacitance and Resistance

For capacitors in series, the reciprocal of the total capacitance is the sum of the reciprocals:\[ \frac{1}{C_{eq}} = \frac{1}{3.8 \times 10^{-6}} + \frac{1}{3.8 \times 10^{-6}} \]For resistors in series, the total resistance is simply the sum:\[ R_{eq} = 2.2 \times 10^3 + 2.2 \times 10^3 = 4.4 \times 10^3 \Omega \].

Step 3: Compute Equations and Values

Calculate the equivalent capacitance:\[ C_{eq} = \frac{3.8 \times 10^{-6}}{2} = 1.9 \times 10^{-6} \mathrm{~F} \]Compute the time constant (\( \tau \)):\[ \tau = R_{eq} \cdot C_{eq} = 4.4 \times 10^3 \cdot 1.9 \times 10^{-6} \approx 8.36 \times 10^{-3}\mathrm{~s} \].

Step 4: Determine Time for Specific Current

The current at time \( t \) in an RC circuit can be found using the equation:\[ I(t) = I_0 \cdot e^{-\frac{t}{\tau}} \]Where \( I_0 \) is the initial current. Solve the equation for the time when \( I(t) = 1.50 \mathrm{~mA} \) and \( I_0 = \frac{12.0}{4.4 \times 10^3} \approx 2.73 \mathrm{~mA} \).

Step 5: Solve for Time

Plug the known values into the equation:\[ 1.50 = 2.73 \cdot e^{-\frac{t}{8.36 \times 10^{-3}}} \]Solving for \( t \),\[ \ln \left( \frac{1.50}{2.73} \right) = -\frac{t}{8.36 \times 10^{-3}} \]\[ t = -8.36 \times 10^{-3} \cdot \ln \left( \frac{1.50}{2.73} \right) \approx 5.12 \times 10^{-3} \mathrm{~s} \].

Key Concepts

Understanding the Time Constant

In RC circuits, the time constant is crucial for determining how fast the circuit responds to changes, such as charging or discharging. It is symbolized by \( \tau \) and is calculated by multiplying the equivalent resistance \( R_{eq} \) by the equivalent capacitance \( C_{eq} \). This product gives the time constant \( \tau = R_{eq} \cdot C_{eq} \) in seconds.
In a practical sense, the time constant gives us the time it takes for the charge or current in a circuit to reach approximately 63% of its final value if starting from zero, or drop exponentially to 37% of its initial value if the circuit is discharging.

• The time constant \( \tau \) provides insight into how quickly a circuit can react.
• RC circuits will take about 5 time constants to fully charge or discharge.

Capacitors in Series

In a series configuration, capacitors are connected end-to-end which affects the total capacitance of the circuit. The formula for calculating the total capacitance \( C_{eq} \) of capacitors in series involves taking the reciprocal of the sum of the reciprocals of the individual capacitance values.
The formula can be expressed as:\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n} \]where \( C_1, C_2, \ldots, C_n \) are the individual capacitances.
Therefore, the equivalent capacitance of capacitors in series is always less than any one of the capacitors' individual capacitance values. This might seem counterintuitive at first, but this arrangement allows the capacitors to work together to store a smaller amount of charge compared to any single capacitor in the series.

• Series connection reduces the overall capacitance.
• Total voltage across series capacitors is the sum of individual voltages.

Resistors in Series

When resistors are connected in series, they are placed end-to-end in a circuit. The total or equivalent resistance \( R_{eq} \) for resistors in series is straightforward to calculate. It is simply the sum of each resistor's individual resistance.
The formula is:\[ R_{eq} = R_1 + R_2 + \cdots + R_n \]where \( R_1, R_2, \ldots, R_n \) are the resistances of each resistor in the series.
This is because each charge passing through the circuit has to pass through each resistor one by one, leading to a sum of all the resistive forces they encounter. As a result, the total resistance in the circuit increases, which can reduce the current flowing through the circuit if the voltage remains constant.

• Resistance adds up in a series circuit.
• The total voltage drop across the series is the sum of each resistor's drop.