184k views
3 votes
A 10 μF (microfarad) capacitor is charged to 100 V and then disconnected. One can model the charge leakage of the capacitor with an RC circuit with no voltage source, representing the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is 4 x 10⁹ ohms; on a humid day, the resistance is 0 x 10⁹ ohms. How long will it take the capacitor voltage to fall to half its original value on each day?

User Figolu
by
9.0k points

1 Answer

4 votes

Final answer:

On a cold dry day, the capacitor's voltage will fall to half its original value in 40 seconds. The situation for a humid day with a resistance of 0 x 10⁹ ohms represents a short circuit, which suggests there is likely a mistake in the question since the resistance cannot realistically be zero.

Step-by-step explanation:

The question is asking about the discharge time of a capacitor in an RC circuit under different environmental conditions. To find how long it takes for the capacitor voltage to fall to half its original value (which is the definition of the RC time constant, τ), we use the formula τ = RC, where R is the resistance and C is the capacitance.

On the dry day, with a resistance of 4 x 10⁹ ohms, the time constant would be: τ = (4 x 10⁹ Ω)(10 x 10⁻⁶ F) = 40 seconds. The voltage falls to half of its initial value exactly after one time constant, which is 40 seconds in this case.

On a humid day, with a resistance supposedly given as 0 x 10⁹ ohms, this situation implies a short circuit, which is not realistic for air resistance. Therefore, the given resistance value on a humid day seems to be incorrect, and the question cannot be answered as posed since resistance cannot be zero. A resistance of zero would imply an instantaneous discharge, which defies real-world conditions and makes the capacitor discharge time negligible.

User Nupadhyaya
by
9.3k points