A 600W flashlamp fitted to a camera is operated by discharging a 10 000 uF capacitor through the lamp. The capacitor is charged initially from a 6V battery.
Calculate
A) the energy stores in the capacitor when fully charged
B) the duration of the flash
watts = volts * amps = amps^2 * Ω
600 W = 6 V * 100 A = 100^2 * Ω
A)energy = 1/2 * farads * (volts)^2
B)use C and R (Ω) to find the time constant of the circuit
the capacitor is 95% discharged after 3 time constants
To calculate the energy stored in the capacitor when fully charged, you can use the formula:
E = 0.5 * C * V^2
where E is the energy stored in joules, C is the capacitance in farads, and V is the voltage across the capacitor.
A) Calculate the energy stored in the capacitor when fully charged:
Given:
Capacitance (C) = 10,000 uF = 10 * 10^-3 F (convert to farads)
Voltage (V) = 6V
Using the formula, substitute the values:
E = 0.5 * (10 * 10^-3) * (6)^2
E = 0.5 * 0.01 * 36
E = 0.18 joules
Therefore, the energy stored in the capacitor when fully charged is 0.18 joules.
B) To calculate the duration of the flash, we can use the formula for energy:
E = P * t
where E is the energy in joules, P is the power in watts, and t is the time in seconds.
Given:
Power (P) = 600W (watts)
Energy (E) = 0.18 joules (from part A)
Rearrange the formula to solve for time (t):
t = E / P
Substitute the values:
t = 0.18 / 600
t = 0.0003 seconds
Therefore, the duration of the flash is 0.0003 seconds or 0.3 milliseconds.