Respuesta :
Answer:
Explanation:
It is a question relating to charging of capacitor . For charging of capacitor , the formula is as follows.
Q = CV ( 1 - [tex]e^{-\lambda\times t}[/tex] )
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 - [tex]e^{-2}[/tex] )
= 2500 x 10⁻⁶ x .86566
= 2161.66 μ C .
voltage = Charge / capacitor
2161.66 μ C / 500μ F
= 4.32 V
Answer:
Voltage across capacitor after 20 sec is 4.323 volt
Explanation:
We have given capacitance [tex]C=500\mu F=500\times 10^{-6}F[/tex]
Battery voltage V = 5 volt
Resistance [tex]R=20kohm=20\times 10^3ohm[/tex]
Time constant of RC circuit
[tex]\tau =RC[/tex]
[tex]\tau =20\times 10^3\times 500\times 10^{-6}=10sec[/tex]
Time is given t = 20 sec
Capacitor voltage at any time is given by
[tex]V_C=V_S(1-e^{\frac{-t}{\tau }})[/tex]
[tex]V_C=5(1-e^{\frac{-20}{10 }})[/tex]
[tex]V_C=5\times 0.864=4.323volt[/tex]
So voltage across capacitor after 20 sec is 4.323 volt
