Question

A 30 μF capacitor is connected across a programmed power supply. During the interval from t = 0 to t = 3.00 s the output voltage of the supply is given by V(t) = 6.00 + 4.00t - 2.00t² volts. At t = 0.500 s

a. find the charge on the capacitor

Answer 1

At t = 0.500 s, the voltage on a 30 µF capacitor is 7.50 V. The charge on the capacitor can be calculated using q = CV, resulting in a charge of 225 µC.

Step-by-step explanation:

To calculate the charge on a 30 µF capacitor at a given time t = 0.500 s, using the provided voltage function V(t) = 6.00 + 4.00t - 2.00t² volts, we first need to plug the time into this function to find the voltage at that specific time.

Substituting t = 0.500 s into the voltage function gives:

V(0.500) = 6.00 + 4.00(0.500) - 2.00(0.500)² = 6.00 + 2.00 - 0.50 = 7.50 V

The charge q on a capacitor is given by the formula q = CV, where C represents the capacitance and V represents the voltage across the capacitor. Using the capacitance value of 30 µF and the voltage at t = 0.500 s:

q = (30 µF) × (7.50 V) = 225 µC

Therefore, the charge on the capacitor at 0.500 s is 225 µC.