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You have a 68.3-mF capacitor initially charged to a potential difference of 11.1 V. You discharge the capacitor through a 3.65-Ω resistor. What is the time constant? After what multiple of the time constant does the potential difference reach 1.83 V?

Respuesta :

Answer:

0.25 sec

1.8 times

Explanation:

C = Capacitance of the capacitor = 68.3 mF = 68.3 x 10⁻³ F

R = resistance of the resistor = 3.65 Ω

V₀ = Maximum potential difference to which the capacitor is charged = 11.1 Volts

Time constant is given as

T = RC

T = (3.65) (68.3 x 10⁻³)

T = 0.25 sec

V = final potential difference = 1.83 Volts

Using the equation

[tex]V = V_{o}(e^{\frac{-t}{T}})[/tex]

[tex]1.83 = 11.1( e^{\frac{-t}{T}})[/tex]

[tex]\frac{t}{T} = 1.8[/tex]

t = 1.8 T

So 1.8 times

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