Question

4. An inductor of inductance L= 15 H is charged through a resistor of resistance R=5 W by a battery of voltage V= 20 Volts. The current as a function of time during charging of this inductor is represented in the table below. The equation of current versus time is also represented by I(t)=L(1 - exp(-t/t) ), where L =V/R, and t = L/R. By taking In of both sides of the current equation, one could linearize it as: In(I-I(t)) = ln(L)- t/t. t(second) 0 1 2 3 4 5 01.13 1.95 2.53 2.95 3.24 -I) I(A) Ins (a) Complete the last row of the table above and using excel plot In(I.-I) vs t and display the equation of best fitted line and its slope. (b) Using the slope in part (a) and the relation t = - 1/slope, calculate the time constant t. (c) Using t =L/R, calculate the theoretical time constant t. (d) From the data table, calculate I(t)/I.

Answer 1

Time (t) second l(t) A In[Imax - I(t)] Long Name Units Comments F(x)= 1 0 Imax - I(t) A Imax = 4A 4-col(B) 4 2.87 2.05 1.47 1.05 0.76 1 2 3 0 1.13 1.95 2.53 2.95 3.24 WN In(col(C)) 1.38629 1.05431 0.71784 0.38526 0.04879 -0.27444 4 5 6 4 5

* In[Imax - |(t)] Linear Fit of Sheet1 D'In[Imax - 1(t)]"| 1.6 1.4 1.2 1.0 Equation Plot Weight Intercept Slope Residual Sum of Squares Pearson's r R-Square (COD) Adj. R-Square y = a + b*x In[Imax - l(t)] No Weighting 1.38535 + 0.00271 -0.33294 + 8.94033E-4 5.59506E-5 -0.99999 0.99997 0.99996 0.8 0.6 In[Imax - l(t)] 0.4 0.2 0.0 -0.2 -0.4 0 1 2 3 4 5 Time (t) (second)

4.) ) L= 15 H ; V=20 Volts, R=5w Cit should be a notw) : Imax = (1/²) = 4A now, put the data in origin lab and calculate bin (Iman - Ilt)) and plot it with, time. After fitting linearely we get (slope 0.33294 S- (6) : Time constant it L slope =3 sec Cuppron) (3:00354s (2) Theomitical time constant, It - - - 3 sec (d) I (t) = Iman Ple 1 (t) Iman (1- exp(+/+)) 1-ero (+/ froom data table at any t, Ilt) given, so, I) I(t) Iman 3 4