Ch 28: Fundamentals of Circuits

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 28: Fundamentals of Circuits

Problem 72

Chapter 28, Problem 72

A 15 μF capacitor charged to 12 V is discharged through a resistor. The energy stored in the capacitor decreases by 50% in 0.25 s. What is the value of the resistance?

Verified step by step guidance

Step 1: Start by recalling the formula for the energy stored in a capacitor: \( E = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the voltage across the capacitor. This will help us understand how the energy changes over time.

Step 2: Use the fact that the energy decreases by 50% in 0.25 seconds. The energy at any time \( t \) during discharge is given by \( E(t) = E_0 e^{-\frac{t}{RC}} \), where \( E_0 \) is the initial energy, \( R \) is the resistance, and \( C \) is the capacitance. Substitute \( E(t) = 0.5 E_0 \) and \( t = 0.25 \) s into this equation.

Step 3: Simplify the equation \( 0.5 E_0 = E_0 e^{-\frac{0.25}{RC}} \). Cancel \( E_0 \) on both sides to get \( 0.5 = e^{-\frac{0.25}{RC}} \). Take the natural logarithm of both sides to isolate \( \frac{0.25}{RC} \): \( \ln(0.5) = -\frac{0.25}{RC} \).

Step 4: Rearrange the equation \( \ln(0.5) = -\frac{0.25}{RC} \) to solve for \( R \): \( R = -\frac{0.25}{C \ln(0.5)} \). Substitute \( C = 15 \mu F = 15 \times 10^{-6} \) F into the equation.

Step 5: Perform the substitution and simplify the expression for \( R \). Ensure the units are consistent throughout the calculation. The resistance \( R \) can now be calculated using the given values.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is the ability of a capacitor to store electrical energy in an electric field, measured in farads (F). It is defined as the charge stored per unit voltage across the capacitor. In this case, the 15 μF capacitor can store energy based on its capacitance and the voltage applied, which is crucial for understanding how much energy is initially stored.

Energy Stored in a Capacitor

The energy (E) stored in a capacitor can be calculated using the formula E = 1/2 CV², where C is the capacitance and V is the voltage. This relationship is essential for determining how much energy is present in the capacitor before and after discharge. In this scenario, knowing that the energy decreases by 50% allows us to calculate the initial and final energy values.

RC Time Constant

The RC time constant (τ) is a measure of the time it takes for the voltage across a capacitor to charge or discharge through a resistor. It is calculated as τ = R × C, where R is the resistance and C is the capacitance. This concept is vital for understanding the rate of energy loss in the circuit, particularly since the problem states that the energy decreases by 50% in a specific time frame.

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