Variation of Voltage Across a Capacitor
See attached file for full problem description. Please answer only question 1.2
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SOLUTION
Volts
I(t)
6
+
150nF V(t) 4 5V

2
Î¼sec
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 Î¼sec : The voltage is constant at 4 V. Hence, dV/dt = 0. From (1), I(t) = 0
ii) t = 2 Î¼sec to t = 4 Î¼sec : Change in voltage = 4 V and change in time = 6Î¼sec = 6x106 sec
dV/dt = 4/(6x106)
From (1) I(t) = 150 x 109 x 4/(6x106) = 0.1 A
iii) t = 4 Î¼sec to t = 5 Î¼sec : 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.