A 1.80 mH inductor is connected in series with a dcbattery of negligible interna
ID: 1721744 • Letter: A
Question
A 1.80 mH inductor is connected in series with a dcbattery of negligible internal resistance, a 0.900 kresistor, and an open switch. How long after the switch is closed will ittake for the current in the circuit to reach half of its maximumvalue ? also.. How long after the switch is closed will ittake for the energy stored in the inductor to reach half of itsmaximum value? A 1.80 mH inductor is connected in series with a dcbattery of negligible internal resistance, a 0.900 kresistor, and an open switch. How long after the switch is closed will ittake for the current in the circuit to reach half of its maximumvalue ? How long after the switch is closed will ittake for the current in the circuit to reach half of its maximumvalue ? also.. How long after the switch is closed will ittake for the energy stored in the inductor to reach half of itsmaximum value? How long after the switch is closed will ittake for the energy stored in the inductor to reach half of itsmaximum value?Explanation / Answer
. 01. Imax = V/R 02. I = (Imax)*{1-exp[(-t)/()] } ; equationfor the time-varying current in a "charging" RL circuit 03. = time constant =L/R 04. U = energy stored in an inductor =(1/2)*(L)*(I2) . 05. = (1.8E-3 Henry)/(0.900E+3 ) =2.00E-6 seconds 06. Part a: "How long after the switch isclosed will it take for the current in the RL circuit to reach halfof its max value?" 07. (I)/(Imax) =0.5 ; given inproblem statement 08. 0.5 = 1 - exp[(-t)/(2.00E-6seconds)] ; takethe natural logrithm of both sides 09. ln(0.5) = (-t)/(2.00E-6seconds) ; solvefor t 10. t = (-2.00E-6seconds)*[ln(0.5)] = 1.386294E-6 seconds . 11. Part b: "How long after the switch isclosed will it take for the energy stored in the inductor to reachhalf of its max value?" 12. Square both sides of'02' 13. (I2) =(Imax)2*{1-exp[(-2t)/()]} 14. Multiply both side of '13'by (L/2) 15. (L/2)*(I2) =(L/2)*(Imax)2*{1-exp[(-2t)/()]} 16. Note that '15' representsthe time varying energy within the inductor (compare with'03') 17. Using '15'... 18. [(L/2)*(I2)] / [(L/2)*(Imax)2]= 0.5 ; givenin problem statement 19. 0.5 = 1 -exp[(-2t)/()] ; takethe natural logrithm of both sides 20. ln(0.5) =(-2t)/() ; solvefor t 21. t =(1/2)*(-2.00E-6 seconds)*[ln(0.5)] = 693.1472E-9 seconds .Related Questions
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