Consider the circuit in the figure below. Take = 6.00 V, L = 6.00 mH, and R = 6.
ID: 2012549 • Letter: C
Question
Consider the circuit in the figure below. Take = 6.00 V, L = 6.00 mH, and R = 6.00 .
(a) What is the inductive time constant of the circuit? ___?___ ms
(b) Calculate the current in the circuit 250 µs after the switch is closed. ___?___ A
(c) What is the value of the final steady-state current? ___?___ A
(d) How long does it take the current to reach 80% of its maximum value? ___?___ ms
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Explanation / Answer
a) Inductive Time constant = (L/R) = 6mH/6 = 1 m sec
b) Inductor behaves as an open circuit just after closing the switch, and behaves as a short circuit at steady state (at infinite time). Hence the Current in RL series circuit for a step voltage will be initially "0" and reaches maximum upto (V/R) when Inductor becomes short circuit. The time to reach the maximum current depends upon time constant , i.e., L/R
Solving a linear differential equation for RL ckt, di/dt + (R/L) i(t) = V(t),
i(t) = V/R [ 1- e(-R/L) t]
hence The current after 250 s will be, i(t)= 6/6 [ 1- e-(6/6m)250] = 1[1-e-0.25] = 0.221 A
c)The value of final steady state current is V/R = 6/6 = 1A
d) to reach 80% of maximum current, i.e., 0.8 A,
0.8 = (1- e-1000 t)
e-1000 t =0.2
e1000 t=5
t = ln 5 m sec
t= 1.609 ms
hence The current reaches 80% of its maximum value after 1.609 m sec