A V=108V source is connected in series with an R=1.3 kOhms resistor and an L=34H

Question

A V=108V source is connected in series with an R=1.3 kOhms resistor and an L=34H 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. After the current decreases to 48% of its maximum value, the battery is reconnected into the circuit by reversing the switch. At 42.91 milliseconds the current reaches 79% of its maximum.

How much energy, in millijoules, is supplied in total by the battery, both in initially bringing the current to maximum and in bringing the current back to the 79% level from 48%? (ignore energy dissipated in the resistor during those processes)

189mJ IS INCORRECT!

Explanation / Answer

given V = 108 v
R = 1300 ohm
L = 34 H

so maximum current = Io
Io = V/R = 108/1300 = 83.0769 mA
energy supplied by battery to the inductor = 0.5*L*i^2 = 117.33 mJ
now, eenrgy supplied to being curernt from 48% to 79% = 0.5*LIo^2(0.79^2 - 0.48^2) = 46.1928 mJ
total energy = 163.5228 mJ

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