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

Answer:

Explanation:

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

Q = CV ( 1 - $e^{-\lambda\times t}$ )

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

Answer 2

Answer:

Voltage across capacitor after 20 sec is 4.323 volt

Explanation:

We have given capacitance $C=500\mu F=500\times 10^{-6}F$

Battery voltage V = 5 volt

Resistance $R=20kohm=20\times 10^3ohm$

Time constant of RC circuit

$\tau =RC$

$\tau =20\times 10^3\times 500\times 10^{-6}=10sec$

Time is given t = 20 sec

Capacitor voltage at any time is given by

$V_C=V_S(1-e^{\frac{-t}{\tau }})$

$V_C=5(1-e^{\frac{-20}{10 }})$

$V_C=5\times 0.864=4.323volt$

So voltage across capacitor after 20 sec is 4.323 volt