can someone help me with part b please Part A A240 mH inductor is connected in s

Question

can someone help me with part b please

Part A A240 mH inductor is connected in series with a dc battery of negligible internal resistance, a 0 840 k resistor, and an open switch. How long after the switch is closed will it take for the current in the circuit to reach half of its maximum value? tl. 1.98 Submit My Answers Give Up Correct Part B How long after the switch is closed will it take for the energy stored in the inductor to reach half of its maximum value? Hints Hint 1 You will have to solve an exponential equation. t2 . 1.99 Submit MyAnswer GiveUp Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

According to the given problem,

L = 2.4 mH
R = 840

This will produce a very short time constant T = L / R = 3.86 * 10-6 s

The current is

i = Imax * (1 - e-t/T)

When i = 0.5*Imax
0.5 = 1 - e-t/T  
t = 1.98*10-6 s = 1.98 s


b)Emax = 0.5*L*Imax2

When E = 0.5*Emax
i2 / Imax2 = 0.5
so i = Imax / 2 = 0.7071*Imax

0.7071 = 1 - e^(-t/T)  
t = 3.508*10-6 s = 3.51 s

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