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Postby TGTS0907129» Find induced emf at steady current value

A coil of resistance 3.0O and inductance 0.25 H is connected to a battery of negligible internal resistance and terminal potential difference 60 V. what is the induced emf at the instant the current has risen to one-fourth its final steady value? At what is the current changing at this instant? What is the time constant for this circuit


The steady current is given by
I = V/R = (60 V)/3.0 = 20A

The instantaneous current is 20 A/4 = 5.0 A,
V = iR + vL
vL = V - iR = 60V - (5.0 A × 3.0Ω) = 45 V
?i/?t = vL/L = (45 V)/0.25H = 180 A/s

The time constant is
L/R = (0.25 H)/3.0Ω = 0.083s

If after the current has reached the steady value, the switch S is thrown quickly from position 1 to position 2, the current will be die down exponentially. In the new circuit V = 0m, and during the decay
0 = iR + vL

In the decay the current decrease and the induced emf is in such a direction as to resist this change. The induced emf is thus a forward emf. The change direction remains the same as before, but vL reverses. Note that during the rise the ordinates of the two curves add al all times to V; during the decay they add to zero.

By TGTT170912123 on 9/30/2015 12:48:28 AM
TGTS0907129 on 9/29/2015 11:04:52 PM

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