At time t = 0 ms, a 450 V potential difference is suddenly applied to a coil with L = 50.0 mH and R = 180ohm .At what rate is the current increasing at t = 1.20 ms?
In LR Circuit Current is given by:
i = (E/R)*(1 - exp(-Rt/L))
Rate of current increase will be given by:
di/dt = (E/R)*(0 - (-R/L)*exp(-Rt/L))
di/dt = (E/L)*exp(-Rt/L)
Now given values are:
E = 450 V
L = 50 mH = 0.05 H
R = 180 Ohm
At t = 1.20 ms = 1.20*10^-3 sec
Current increasing rate will be:
di/dt = (450/0.05)*exp(-180*1.20*10^-3/0.05)
di/dt = 119.7 Amp/sec
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Also Check that generally Voltage is very low than 450 V, So please check the given value of V in your question, is it 450 V OR 45.0 V
If it's 45.0 V, then your answer will be di/dt = 11.97 Amp/sec = 12 Amp/sec
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