Question

A 25 μF capecitor charged to 12 V is discharged through a Part A resiator. The energy stored in the capacitor decreases by 50% in 0.25 s. What is the value of the resistance?

Answer 1

The resistance through which a 25 μF capacitor discharges, resulting in a 50% energy decrease after 0.25 s, is calculated using the time constant formula and found to be 10,000 ohms or 10 kΩ.

Step-by-step explanation:

To find the value of the resistance in a circuit where a 25 μF capacitor charged to 12 V is discharged through a resistor and the energy stored in the capacitor decreases by 50% in 0.25 s, we can use the following relationship between the time constant (τ), resistance (R), and capacitance (C):

τ = R × C

The time constant τ is the time it takes for the voltage (or charge) to decay to about 36.8% of its initial value. When the energy stored in the capacitor decreases by 50%, this corresponds to one time constant because the energy stored in the capacitor, which is proportional to the square of the voltage (or charge), decreases to 50% when the voltage decreases to 70.7% (square root of 50%) of its initial value. Hence, we are given the time constant τ is 0.25 s, and the capacitance C is 25 μF (or 25 x
`10^{-6` F).

Using the relationship:

τ = R × C

R = τ / C

R = 0.25 s / (25 x
`10^{-6` F)

R = 10,000 Ω

The value of the resistance is 10,000 ohms or 10 kΩ.