# Video: Calculating the Properties of an RC Circuit

The duration of a photographic flash is related to an ππΆ time constant, which is 0.100 πs for a certain camera. The capacitor in the camera is in a circuit with a charging resistance of 800 kΞ©, while the flash lamp of the camera has a resistance of 40.0 mΞ© during discharge. What is the capacitance of the capacitor? What is the time constant for charging the capacitor?

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### Video Transcript

The duration of a photographic flash is related to an ππΆ 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πΆ circuitβs time constant 0.100 microseconds, which weβll 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 weβll call π sub π.

In part one, we wanna solve for the capacitance of the capacitor. Weβll call that capital πΆ. And in part two, we wanna solve for the time constant not for discharging, but for charging a capacitor β what weβll π sub π.

We can start off solving this problem by recalling the definition for a time constant for an ππΆ circuit. In a ππΆ circuit, the time constant, which we can call π, is equal to the product of the circuitβs 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 πΆ, the capacitance, or πΆ equals π over π sub π.

Weβre told π and π sub π in our problem statement. So we can plug in and solve for πΆ. When we do, weβre 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 πΆ is 2.50 times 10 to the negative six farads or 2.50 microfarads. Thatβs 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 itβs charging rather than discharging. Weβll again use our time constant relationship. But this time instead of using π sub π, weβll use π sub π β the effective resistance of our circuit when it charges.

Weβre given π sub π in the statement and weβve solved for πΆ, the capacitance of our circuit, earlier. So weβre ready to plug in and solve for π sub π. When we do, weβre 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. Thatβs the time constant of the circuit while it charges.