A 12.6 V battery is in series with a 30.0 mH inductor and 0.150 ohm resistor connected through a switch. When the switch is closed at t o, find time constant of the circuit( ans: 0.2 s) Find the current after 1 time constant has elapsed ( ans: 53.1 A) Find the voltage drop across R after t0 and after one time constant ( ans: 0 volts and 7.97 V ) find rate of change of current after one time constant( ans: 150 A/s)I have the answers, but I need a step by step? I'd appreciate any help.
Let V_s the voltage of the battery 12.6 V Let i the current through the series circuit Let R the resistance of the resistor 0.150 Ω Let L the inductance of the inductor 0.03 H Let V_r the voltage across the resistor (i)R Let V_l the voltage across the inductor L(di/dt) The source voltage must equal the sum of the voltages across the components: V_s V_r + V_l 12.6 V (i)R + L(di/dt) di/dt + (i)(R/L) (12.6 V)/L The integrating factor for this is e^{∫ (R/L)dt} e^{(R/L)t} e^{(R/L)t}di/dt + e^{(R/L)t}(i)(R/L) e^{(R/L)t}(12.6 V)/L The left side integrates as the reverse of the product rule and the right side integrates with the reciprocal of the coefficient with a constant, C: e^{(R/L)t}(i) e^{(R/L)t}(12.6 V)/R + C Multiply both sides by e^{-(R/L)t}: (i) (12.6 V)/R + Ce^{-(R/L)t} i (12.6 V)/0.150 Ω + Ce^{-(R/L)t} i 84 A + Ce^{-(R/L)t} We find the value of C by knowing that i 0 at t 0 0 84 A + Ce^0 C - 84 A i (84 A)(1 - e^{-(R/L)t}) To find the time constant set (R/L)t 1: t L/R 0.03/0.150 0.2 s One time constant means that -(R/L)t -1 i (84 A)(1 - e^-1) ≈ 53.1 A The current is 0 at t 0 so V_r R(0) 0 The current is 53.1 A at t 0.2 s so V_r (0.150 Ω)(53.1 A) ≈ 7.97 V The charge rate is di/dt and we have an equation involving that: di/dt + (i)(R/L) (12.6 V)/L Solve for di/dt: di/dt (12.6 V)/L - (i)(R/L) di/dt 12.6 V/0.03 H - (53.1 A)(0.150 Ω/0.03 H) di/dt 154.5 A/s
