In an RL circuit connected to a 12 V battery at time t = 0, the current is measu

Question

In an RL circuit connected to a 12 V battery at time t = 0, the current is measured to be 0.2 A after 1.4 10-4 s, and 0.34 A after 10 s. What are the values of R and L?
R____?__ohms
L___?___ mH


starting it out....

i(t) = i (1-e^-(R/L)t)
0.2 = i (1 - e^-(R/L)0.0001.4) ----------------1
0.34 = i (1-e^-(R/L)10)-----------------2
two unknown and two equation
dividing eqtn 1 from 2 and taking logarithm we can get the answers. but howww???

Explanation / Answer

Here, actually we have not 2 variables but 3 variables, viz. R,L,i=final value= V/R i(t) = i (1-e^-(R/L)t) 0.2 = i (1 - e^-(R/L)0.00014) ----------------1 0.34 = i (1-e^-(R/L)10)-----------------2 dividing the two equations,(eq (1)/eq(2)) we get 0.588 = (1-e^-(R/L)*0.00014)/(1-e^-(R/L)*10) taking natural logarithm in both sides, we get, ln(0.588) = ln( (1-e^-(R/L)*0.00014)/(1-e^-(R/L)*10)) so, -0.531 = ln( (1-e^-(R/L)*0.00014)) - ln(1-e^-(R/L)*10) ------> (using the property, ln(a/b) = lna -lnb) so, -0.531 = (R/L)*.00014 - (R/L)*10 so, so, R/L = 0.0531 also, i = final value = V/R = 12/R from equation (1), 0.588 = 12/R(1 - e^-(R/L)0.00014) = 12/R(1-e^-7.434*10^-6) taking natural logarithms both sides, -0.531 = ln(12/R) + 7.434*10^-6 so, taking antilog of both sides, we get, R = 21.41 ohms

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