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A 20 μF capacitor has previously been charged up to contain a total charge of Q=100 μC on it. The capacitor is then discharged by connecting it directly across a 100kΩ resistor. Calculate the charge remaining on the capacitor exactly 3.00 seconds after being connected to the resistor.

Answer :

Answer:

Q= 22.3 μC

Explanation:

Given that

C= 20 μF

Qo= 100 μC

R= 100 kΩ

t= 3 s

T= R C

T= 100 x 1000 x 20 x 10⁻⁶ s

T=2 s

We know that charge on the capacitor is given as

[tex]Q=Q_0e^{\dfrac{-t}{T}}[/tex]

[tex]Q=100\times 10^{-6}\times e^{\dfrac{-3}{2}}[/tex]

Q= 0.0000223 C

Q= 22.3 μC

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