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# Variation of Voltage Across a Capacitor

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See attached file for full problem description. Please answer only question 1.2

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#### Solution Preview

Please see attached file.

SOLUTION

Volts
I(t)
6
+
150nF V(t) 4 5V
-
2
μsec
-4 -2 2 4 6
-4V

a) Charge Q on a capacitor, its capacitance C and the voltage V across it are related as :

Q = CV

Current flowing through the capacitor is equal to the rate of change of charge Q.

I(t) = dQ/dt = C dV/dt .............(1)

Thus, current flowing during different time intervals is calculated using (1) as follows :

i) Upto t = -2 μsec : The voltage is constant at -4 V. Hence, dV/dt = 0. From (1), I(t) = 0

ii) t = -2 μsec to t = 4 μsec : Change in voltage = 4 V and change in time = 6μsec = 6x10-6 sec

dV/dt = 4/(6x10-6)

From (1) I(t) = 150 x 10-9 x 4/(6x10-6) = 0.1 A

iii) t = 4 μsec to t = 5 μsec : Change in voltage = 5 ...

#### Solution Summary

Step by step solution provided which includes graphical representation of the current through the capacitor and power flowing through it.

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