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iragen [17]
3 years ago
5

A 8.0 µF capacitor is initially connected to a 2.0 V battery. Once the capacitor is fully charged the battery is removed and a 6

.0 Ω resistor is connected between the two terminals of the capacitor. Find the current flowing through the resistor 13.0 µsec after the capacitor begins to be discharged.
Physics
1 answer:
aalyn [17]3 years ago
3 0

Answer:

I = 0.451 amp

Explanation:

given,

C = 8.0 µF

V = 2 V

resistor connected between two terminal = 6 Ω

current flowing through resistor = 13 µsec

Q = CV

Q = 8 x 2

Q = 16 µC                              

for an RC discharge circuit

V = V_0e^{-\dfrac{t}{RC}}

I = \dfrac{-Q_0}{RC}e^{-\dfrac{t}{RC}}

t =  13 µsec

I = \dfrac{-16}{6\times 2}e^{-\dfrac{13}{6\times 2}}

I = 0.451 amp

neglecting  -ve sign just to show direction.

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