See attached file for full problem description. Please answer only question 1.2© BrainMass Inc. brainmass.com October 24, 2018, 8:59 pm ad1c9bdddf
Please see attached file.
150nF V(t) 4 5V
-4 -2 2 4 6
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 ...
Step by step solution provided which includes graphical representation of the current through the capacitor and power flowing through it.
RC Circuit: Potential and Current Variation
In the circuit shown in the figure (see attachment) both capacitors are initially charged to 40.0 V.
1) How long after closing the switch S will the potential across each capacitor be reduced to 10.0 V?
2) What will be the current at that time?View Full Posting Details