A 500 μF capacitor is wired in series with a 5 V battery and a 20 kΩ resistor. What is the voltage across the capacitor after 20 seconds of closing the circuit? Show all calculations in your answer.

Explanation:

It is a question relating to charging of capacitor . For charging of capacitor , the formula is as follows.

Q = CV ( 1 - )

λ = 1/CR , C is capacitance and R is resistance.

= 1/(500 x 10⁻⁶ x 20 x 10³ )

= .1

λ t = .1 x 20

λ t = 2

CV = 500 X 10⁻⁶ X 5

= 2500 X 10⁻⁶ C

Q = 2500 x 10⁻⁶ ( 1 - )

= 2500 x 10⁻⁶ x .86566

= 2161.66 μ C .

voltage = Charge / capacitor

2161.66 μ C  / 500μ F

= 4.32 V

trig 15.3 cos