Past Paper 2002 - Q9
Circuit with battery EMF 12V, a switch, a resistor R and an inductor L.
There is a curved graph of I/t... the usual.
(a) State the magnitude of back emf at the instant the switch is closed <--- how?
0
Back emf question.
Started by Vixus, Jan 29 2006 12:47 PM
2 replies to this topic
#2
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Posted 29 January 2006 - 05:19 PM
I just did this question like 10min ago and think the answer is 12V as at the instant the switch is closed the current in the resistor is zero. This means the back e.m.f must be equal in magnitude to E. At that instant dI/dt is at its greatest, so by
e=-L dI/dt the induced back emf is at its greatest at that point and infact is equal to the supply voltage. Thats what I thought anyway.
e=-L dI/dt the induced back emf is at its greatest at that point and infact is equal to the supply voltage. Thats what I thought anyway.
Nothing!!!!!!!!
#3
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Posted 11 February 2006 - 10:24 PM
I found out that the maximum e that can be achieved is equal to the source e.m.f... so yeah, thanks.
emax = Vemf
emax = Vemf
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