The capacitor that powers the flash in a camera is charged using ^.0V 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?
its 6v battery.
To find the capacitance of the capacitor, we can use the formula:
C = (2 * P * t^2) / V^2
Where:
- C is the capacitance (in Farads)
- P is the power (in Watts)
- t is the discharge time (in seconds)
- V is the battery voltage (in volts)
In this case, we are given:
- P = 20 W
- t = 1.0 μs = 1.0 x 10^(-6) s
- V = ^.0 V
Let's plug in these values and calculate the capacitance:
C = (2 * 20 * (1.0 x 10^(-6))^2) / ^.0^2
First, let's calculate the denominator:
V^2 = ^.0^2 = ^.0^2 = ^.0
Now, let's calculate the numerator:
(2 * 20 * (1.0 x 10^(-6))^2) = 2 * 20 * (1.0 x 10^(-12)) = 4.0 x 10^(-11)
Finally, let's divide the numerator by the denominator to find the capacitance:
C = (4.0 x 10^(-11)) / ^.0 = 4.0 x 10^(-11) F
Therefore, the capacitance of the capacitor is 4.0 x 10^(-11) Farads.