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An LR circuit contains an inductor of 500 mH, a resistor of 25.0 Ω and an emf of 5.00 V in series. Find the potential difference across the resistor at t = (a) 20.0 ms, (b) 100 ms and (c) 1.00 s.
3 years ago

Kevin Nash
332 Points
Sol. L = 500 mH, R = 25 Ω, E = 5 V

a) t = 20 ms
i = i0 (1 – e^–tR/L) = E/R (1-E^-tR/L)
= 5/25 (1 – e ^-20x10-3 x 25/100x 10^-3) = 1/5 (1-e^-1)
= 1/5 (1-0.03678) = 0.1264
Potential difference iR = 0.1264 × 25 = 3.1606 V = 3.16 V.
b) t = 100 m
i = i base 0 (1 – e^-tR/L) = E/R (1 – E^-tR/L)
= 5/25 (1 – e ^-100x10^-3 x 20/100x 10^-3 ) = 1/5 (1 – e^-5)
= 1/5 (1- 0.0067) = 0.19864
Potential difference = iR = 0.19864 × 25 = 4.9665 = 4.97 V.
c) t = 1 sec
i = i base 0 ( 1 – e^-tR/L) = E/R (1 – E^-tR/L)
= 5/25 (1 – e ^-1x25/100/10^-3) = 1/5 (1 – e^-50)
= 1/5 x 1 = 1/5 A
Potential difference = iR = (1/5 × 25) V = 5 V.
3 years ago
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