### Video Transcript

The duration of a photographic flash is related to an π
πΆ time constant, which is 0.100 microseconds for a certain camera. The capacitor in the camera is in a circuit with a charging resistance of 800 kiloohms, while the flash lamp of the camera has a resistance of 40.0 milliohms during discharge. What is the capacitance of the capacitor? What is the time constant for charging the capacitor?

This statement tells us this π
πΆ circuitβs time constant 0.100 microseconds, which weβll call π. It also tells us that when the circuit is charging, it has a resistance of 800 kiloohms β a resistance value we will name π
sub π. And that while the circuit is discharging, its resistance is 40.0 milliohms β what weβll call π
sub π.

In part one, we wanna solve for the capacitance of the capacitor. Weβll call that capital πΆ. And in part two, we wanna solve for the time constant not for discharging, but for charging a capacitor β what weβll π sub π.

We can start off solving this problem by recalling the definition for a time constant for an π
πΆ circuit. In a π
πΆ circuit, the time constant, which we can call π, is equal to the product of the circuitβs resistance times its capacitance. In our case, we can write that π is equal to π
sub π, the discharge resistance in our circuit, multiplied by πΆ, the capacitance, or πΆ equals π over π
sub π.

Weβre told π and π
sub π in our problem statement. So we can plug in and solve for πΆ. When we do, weβre careful to use units of seconds for our time constant and units of ohms for our resistance. When we calculate this fraction, we find that πΆ is 2.50 times 10 to the negative six farads or 2.50 microfarads. Thatβs the capacitance in our circuit whether charging or discharging.

Next, we wanna solve for π sub π, which is the time constant of our circuit when itβs charging rather than discharging. Weβll again use our time constant relationship. But this time instead of using π
sub π, weβll use π
sub π β the effective resistance of our circuit when it charges.

Weβre given π
sub π in the statement and weβve solved for πΆ, the capacitance of our circuit, earlier. So weβre ready to plug in and solve for π sub π. When we do, weβre sure to use units of ohms for resistance and farads for capacitance. Calculating this product, we find that π sub π is equal to 2.00 seconds. Thatβs the time constant of the circuit while it charges.