CapacitanceHard
Question
The charge on each of the capacitors 0.16 ms after the switch S is closed in figure is :
Options
A.24 μC
B.26.8 μC
C.25.2 μC
D.40 μC
Solution
q = 
q = 40(1 − e−1) μC = 40 (1 − 0.37) = 25.2μC
q = 40(1 − e−1) μC = 40 (1 − 0.37) = 25.2μC
Create a free account to view solution
View Solution FreeMore Capacitance Questions
In the given circuit, with steady current, find the potential drop across the capacitor....A capacitor of 0.2 μF capacitance is charged to 600 V. On removing the battery, it is connected with a 1.0 μF ...In the given circuit at t = 0 switch is on. Then after how much time voltage across resistance is 37 volt...In the given circuit, with steady state, the potential difference across the capacitor must be :-...A capacitor of capacitance C0 is charged to a potential V0 and then isolated. A capacitor C is then charged from C0, dis...