The current in the RL circuit shown below reaches half its maximum value in 2.8 ms after the switch S1 is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if L=650 mH.
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Step 1 Given data I = I0* ( 1-e- R/L*t )According to the question the value of I = I0 /2 t = 2.8 ms L = 350 mH Step 2 time constant calculation 1/2 * I0=I0 * ( 1 - e- R/ L *2.8 *10-3 )1 /2 = ( 1 - e - R/L *2.8*10^-3 )e - R / L * 2.8*10-3 = 1- 1/2 e- R / L * 2.8*10-3 = 1/2 taking log on both sides  ln  e- R / L * 2.8 *10-3  = ln 1/2 - R / L * 2.8*10-3  lnee= ln 1/2 We know that ln ee = 1R  / L  * 2.8 * 10-3 = 0.693  Time constant , ( L / R ) = 2.8*10^-3 / 0.6931  L / R = 4.03 * 10^-3 seconds L / R = 4.03 ms Step 3 The resistance is The resistance R = 650 * 10^-3 / 4.30 * 10^-3 = 161.29 Ω ...