A V = 118-V source is connected in series with an R = 1.5-k ohm resistor and an

Question

A V = 118-V source is connected in series with an R = 1.5-k ohm resistor and an L = 26-H inductor and the current is allowed to reach maximum. At time t=0 a switch is thrown that disconnects the voltage source, but leaves the resistor and the inductor connected in their own circuit. Randomized Variables V = 118 V R = 1.5 k omega. L = 26 H (a) How much time, in milliseconds, is needed for the current in the circuit to drop to 11% of its value at t = 0? (b) How much energy, in millijoules, does the circuit dissipate during that time?

Explanation / Answer

a)

T = time constant = L/R = 26/1500 = 0.0173 sec

imax = maximum current = V/R = 118/1500 = 0.0787 A

i(t) = imax (e-t/T)

0.11 = e-t/0.0173

t = 0.0382 sec = 38.2 ms

b)

energy dissipated = (0.5) L (i2 - imax2) =  (0.5) L ((0.11 imax)2 - imax2) = (0.5) (26) ((0.11(0.0787))2 - (0.0787)2)

energy dissipated = - 0.0795 J = - 79.5 mJ

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