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In a decaying L-R circuit the current at t = 0 is I. Find the total charge that flows through the resistor during the time it takes for the energy stored in the inductor to drop to one-fourth of its initial value.
- L*I/R
- L*I/(2*R)
- L*I*sqrt(2)/R
- None
Correct answer: L*I/(2*R)
Solution
Inductor energy U = (1/2)L*i². U dropping to U/4 means i² = I²/4, so the current falls from I to I/2. For i(t) = I*exp(-Rt/L), charge q = integral i dt = (L/R)*(i_initial - i_final) = (L/R)*(I - I/2) = (L/R)*(I/2) = L*I/(2R).
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