Question

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?

Answer #2

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**

**Please Upvote.**

**Comment below if you have any query.**

**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**

answered by: anonymous

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