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.