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A (current vs time) graph of the current passing through a solenoid is shown in figure. For which time is the back electromotive force (epsilon) a maximum. If the back emf at t = 3s is e, find the back emf at 1 = 7 s, 15 s and 40 s. OA, AB and BC are straight line segments. |
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Answer» Solution :When rate of change of flux is maximum then induced emf is also t = 10 sec maximum. In graph at t = 5 sec. to slope is maximum, hence rate of change of flux is also maximum and induced emf will also be maximum. For t=0 s to t=5 s , slope `(dI)/(dt)=1/5As^(-1)` Back emf `epsilon=(-LDI)/(dt)` So for t=0 to t=5s (At t=3 s also) Back emf=E=`-L(1/5)=(-L)/5` `therefore e=(-L)/5` ...(1) (i)For 5s `LT t lt 10 s` slope `(dI)/(dt)=(-3)/5 As^(-1)` Back emf for t=5 s to t=10 s, (At t=7s as well) `e_1=(-LdI)/(dt)=-L((-3)/5)` `-3((-L)/5)` `e_1=-3s [ because e=(-L)/5]` (ii) For 10s to `t lt 30` s slope `(dI)/(dt)=2/20=1/10 As^(-1)` Back emf at t=15 s is , `e_2=-L xx (dI)/(dt)` `-LXX(+1/10)` `=(-L)/10=1/2 e [ because e=(-L)/5]` (iii)At `t ge 30s` , slope =0 , So, at t=40 s back emf `e_3=0` Thus at t=7 s induced emf `e_1=-3e` at t=15 s induced emf `e_2=e/2` at t=40 s induced emf `e_3=0` |
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