Question

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.

Answer 1

Voltage across capacitor after 20 sec is 4.323 volt

Step-by-step explanation:

We have given capacitance
`C=500\mu F=500* 10^(-6)F`

Battery voltage V = 5 volt

Resistance
`R=20kohm=20* 10^3ohm`

Time constant of RC circuit

`\tau =RC`

`\tau =20* 10^3* 500* 10^(-6)=10sec`

Time is given t = 20 sec

Capacitor voltage at any time is given by

`V_C=V_S(1-e^{(-t)/(\tau )})`

`V_C=5(1-e^{(-20)/(10 )})`

`V_C=5* 0.864=4.323volt`

So voltage across capacitor after 20 sec is 4.323 volt

Answer 2

Step-by-step explanation:

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

Q = CV ( 1 -
`e^(-\lambda* t)` )

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 -
`e^(-2)` )

= 2500 x 10⁻⁶ x .86566

= 2161.66 μ C .

voltage = Charge / capacitor

2161.66 μ C / 500μ F

= 4.32 V