A 172.0-V battery is connected in series with a 407.0-Ω resistor. According to Faraday’s Law of Induction, cur?
A 172.0-V battery is connected in series with a 407.0-Ω resistor. According to Faraday’s Law of Induction, current can never change instantaneously, so there is always some “stray” inductance. Suppose the stray inductance is 0.165 μH. How long will it take the current to build up to within 0.5% of its final value of 0.423 A after the resistor is connected to the battery?
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Current in the R - L circuit is given by the relation:
I (t) = (V / R) * (1 - e^ - (t / (L/R)) = I(0) * (1 - e^ - (t / (L/R))
(L/R) is called the time constant = (0.165 / 407) * 10^-6 = 4.05 * 10⁻¹⁰
I(0) is the final steady state value (= V/R) and I(0.5%) is 0.5 % 0f I(0)
I(0.5%) / I(0) = 0.005 = 1 - e^ -(t / (L/R))
e^ -(t / (L/R)) = 1 - 0.005 = 0.995
Taking ln.{ e^ -(t / (L/R))} = - (t / (L/R)) = Ln (0.995) = -0.005
t ( I(0.5%) = 0.005*(L/R) = 0.005*4.05*10⁻¹⁰ = 0.5*4.05*10⁻¹² = 2,025 * 10⁻¹² sec.
Therefore time to reach 0.5% of the final current value = 2.025 pico sec.
Time to reach within 0.5% i.e. 99.5% of the final current value = I (99.5%)
I (99.5%) / I (0) = 0.995= 1 - e^ -(t / (L/R))
e^ -(t / (L/R)) = 1 – 0.995 = 0.005
Ln {e^ - (t / (L/R))} = Ln 0.005
- (t / (L/R)) = Ln 0.005 = -5.298
t (I(99.5%) = 5.298*(L/R) = 5.298 * 4.05 * 10⁻¹⁰ sec. or
Time to reach within 0.5% i.e. 99.5% of the final current value = 2.146 nano sec.
( Pl. note that the time is not dependent on the Voltage applied to the circuit if the current is specified as % of the final value)