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Homework Statement
A circuit consists of a capacitor and an inductor. The resistance in the circuit is small and can be neglected. Initially, at t = 0 s, the voltage across the capacitor is at its maximum of VC = 10 V, the charge stored in the capacitor is 1 mC. It is observed that the capacitor discharges to QC = 0 C after t = 0.25 ms. What is the inductance L of the inductor in the circuit?
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam3/fa09/fig8.gif
Homework Equations
Conservation of energy:
Energy Capacitor + Energy Inductor = Qmax^2 / 2C
The Attempt at a Solution
Initial Energy capacitor = .5*(Qmax^2/C) = .005
Final Energy Capacitor = 0
Initial Energy Inductor = 0
Final Energy Inductor = .5*L*I^2
So:
.005 = .5*L*I^2 where I = dQ/dt which is (1E-3)/(.25E-3) = 4
L = 6.25E-4 H which is wrong.
Right answer is 253.3 microH.
What am I doing wrong?
Thanks for any help!
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