A capacitor that powers the flash in a camera is charged using 6.0 V battery. after that capacitor is charged, it is discharged through the flashbulb in just 1.0 microsec. with avg power output of 20 w. whats is the capacitance?
80 W * 10^-6 s is the total discharge energy. That look like 8*10^-5 Joules to me.
Set that equal to the charged capacitor's stored energy ((1/2) C V^2)
and solve for V. They tell you what V = 6.0 V before the discharge.
To find the capacitance of the capacitor, we can use the formula:
C = (2 * P * t^2) / V^2
where:
C = capacitance
P = power output
t = discharge time
V = voltage
Given data:
P = 20 W
t = 1.0 μs = 1.0 * 10^-6 s
V = 6.0 V
Now, let's substitute the given values into the formula:
C = (2 * 20 * (1.0 * 10^-6)^2) / (6.0)^2
Calculating further:
C = (2 * 20 * (1.0 * 10^-6)^2) / (36.0)
C = (2 * 20 * 1.0 * 10^-12) / 36.0
C = (40 * 10^-12) / 36.0
Simplifying:
C = 1.1111 * 10^-12 Farads
Therefore, the capacitance of the capacitor is approximately 1.1111 picofarads (pF).