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.

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.