A 12.6 V battery is in series with a 30.0 mH inductor and 0.150 ohm resistor connected through a switch. When?

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 t=0 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.

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  • 7 years ago
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    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

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