In the circuit shown in the figure Figure 1 S 1 has been cl
In the circuit shown in the figure (Figure 1) , S 1 has been closed for a long enough time so that the current reads a steady 3.35 A Suddenly, S 2 is closed and S 1 is opened at the same instant. Assume that L = 1.9 mH and C = 4.4 ?F.
Part A
What is the maximum charge that the capacitor will receive?
Part B
What is the current in the inductor at this time?
Solution
At steady state, energy stored in the inductor coil is:
E = (1/2)*L*I^2 = (1/2)*(0.0019)*(3.35)^2 = 1.066x10^-2 J
Part A:
Now when the switch S1 is opened and switch S2 is closed -
Then the same charge shall be stored by the capacitor.
=> (1/2)C*V^2 = 1.066x10^-2 J
=> (1/2)*Q^2/C = 1.066x10^-2
=> Q^2 = 2x1.066x10^-2xC = 2x1.066x10^-2x4.4x10^-6 = 9.3808x10^-8
=> Q = 3.063x10^-4 C
Part B:
When the capacitor has maximum charge, current in the inductor = 0 A.
