The flash on a compact camera stores energy in a 120 μF capacitor that is charged to 220 V. When the flash is fired, the capacitor is quickly discharged through a lightbulb with 5.0 Ω of resistance. At what rate is the lightbulb dissipating energy 250 μs after the flash is fired?

Verified step by step guidance

1. Step 1: Understand the problem. The capacitor discharges through the lightbulb, and we need to calculate the rate at which the lightbulb dissipates energy (power) 250 μs after the flash is fired. This involves using the formula for power dissipation in a resistor: \( P = \frac{V^2}{R} \), where \( V \) is the voltage across the resistor and \( R \) is the resistance.
2. Step 2: Recall the formula for the voltage across a discharging capacitor as a function of time: \( V(t) = V_0 e^{-t / \tau} \), where \( V_0 \) is the initial voltage, \( t \) is the time elapsed, and \( \tau \) is the time constant of the circuit. The time constant is given by \( \tau = RC \), where \( R \) is the resistance and \( C \) is the capacitance.
3. Step 3: Calculate the time constant \( \tau \) using \( \tau = RC \). Substitute \( R = 5.0 \ \Omega \) and \( C = 120 \ \mu F = 120 \times 10^{-6} \ \text{F} \). This will give the value of \( \tau \).
4. Step 4: Determine the voltage \( V(t) \) at \( t = 250 \ \mu s = 250 \times 10^{-6} \ \text{s} \) using the formula \( V(t) = V_0 e^{-t / \tau} \). Substitute \( V_0 = 220 \ \text{V} \), \( t = 250 \times 10^{-6} \ \text{s} \), and the calculated \( \tau \) from Step 3.
5. Step 5: Calculate the power dissipated by the lightbulb at \( t = 250 \ \mu s \) using \( P = \frac{V^2}{R} \). Substitute the value of \( V(t) \) from Step 4 and \( R = 5.0 \ \Omega \) into the formula to find the rate of energy dissipation.

Key Concepts

Capacitance

Capacitance is the ability of a capacitor to store electrical energy in an electric field, measured in farads (F). In this scenario, the 120 μF capacitor stores energy when charged to 220 V. The energy (E) stored in a capacitor can be calculated using the formula E = 1/2 C V², where C is capacitance and V is voltage.

Ohm's Law

Ohm's Law relates voltage (V), current (I), and resistance (R) in an electrical circuit, expressed as V = I R. This fundamental principle allows us to determine the current flowing through the lightbulb when the capacitor discharges. Understanding this relationship is crucial for calculating the power dissipated by the lightbulb at any given time.

Power Dissipation

Power dissipation in an electrical component, such as a lightbulb, is the rate at which it converts electrical energy into heat and light, measured in watts (W). It can be calculated using the formula P = I² R, where I is the current through the component and R is its resistance. This concept is essential for determining how much energy the lightbulb dissipates at a specific time after the flash is fired.