In each case below, calculate the inductive time constant, then find the current

Question

In each case below, calculate the inductive time constant, then find the current when t = 2.00 x 10 , Part (a) uses resistance and inductance values that can be set with the sliders, so you can use the animation to verify your calculations. Part (b) uses values outside the slider ranges. In each case, the battery voltage is 10 V. Express your answers in ms and mA. (a) Let R 1.3 and L-1.2 H. =11.5 Your response differs from the correct answer by more than 10%. Double check your calculations. Find I when t 2.00 ms. Xms 1=29 Your response differs from the correct answer by more than 10%. Double check your calculations. (b) Let R-1.3 ks, and L-2.72 H. ms Find 1 when t-2.00 ms 1 mA

Explanation / Answer

part A)

time constaqnt of RL constant is given by T = (L/R)

where L is inductance

R is resistance

T = 1.2/(1.3*100)

T = 9.23 milli secs

Equation of Current is given by I = IO e^-Rt/L

at time t = 2 milli secs

Io = inital Current = V/R

Io = (10/1300)

Io = 7.69 milli Amperes

I = 7.69 mA * e^(-1300*2*10^-3)/(1.2)

I = 0.88 mili AMperes

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part B :

if R = 1300 ohms

L = 2.72 H

then

T = (2.72/1300)

T = 2.1 milli secs

Io = 10/1300 = 7.69 mA

I = 7.69 mA * e^-(1300 * 2*10^-3)/(2.72)

I = 2.95 Milli Amps

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