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Homework Help: Voltage across an Inducter at a specific time in a DC circuit

  1. Nov 15, 2016 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    I(t) = E/R [ 1 - e^-t/tau)] Vseries = Sum of voltages , Vparalel= V1=V2=V3... V=I*R

    3. The attempt at a solution
    File_000 (2).jpeg
  2. jcsd
  3. Nov 16, 2016 #2


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    I don't know what I'm missing here, but you seem to be starting from different values?

    "L=.33H" L= 0.32 H
    "R1=R2=6.5Ω" R1 = R2 = 1.5 Ω

    Then you calculate "Rtot = ? + ? ... = 0.75Ω" and I'm not sure, what Rtot is, how you calculated it.
    From your result, presumably it must be R1 in parallel with R2. I can't see how this can be: the resistors are not in parallel, unless you replace the inductor with a short. Then there would be no voltage across the inductor.

    Then you calculate τ = L/R using l=0.33 and R= 0.75, implying that the 0.33 H inductance is in series with the 0.75Ω resistance.
    As far as I can see, the inductor is not in series with a parallel combination of resistors.

    At this point I stop checking the calculations, as they cannot achieve a correct result.
  4. Nov 16, 2016 #3
    I posted the wrong problem picture.. So the two resistors aren't in paralel when the switch is closed? If YOU were doing this problem how would you start it?
  5. Nov 16, 2016 #4


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    I think I'd get in trouble if I told you this before you've sorted it out by your own way.

    The top of R1 is joined to the top of R2, but the bottom of R1 is not connected to the bottom of R2, (whether the switch is open or closed), so I'd say they are not in parallel.

    Ok. So we're working from the circuit diagram in the question?

    Your equation, I(t) = E/R [ 1 - e^-t/tau)] , what is it the formula for?
    Well you know it's a current at a time t (I can see that from your working), but what current? And what sort of circuit does it apply to? And what are the E and R values in relation to that circuit?
  6. Nov 16, 2016 #5
    I I re did my calculations and found I(t) for the circuit at time stated. I then used Kirchoffs loop rule for the outter loop : 12 V - R2*I(t) -L di/dt = 0

    So 12 V - R2* I(t) = L di/dt = V(L) which gave me the correct answer of 9.8 Volt.
  7. Nov 16, 2016 #6
    BTW I had to use Resistance in parallel for the I(t) equation.
  8. Nov 16, 2016 #7


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    Fine. You got there on your own.
    Obviously I can't comment on your calculations without seeing them.

    Starting with L=0.32 H, I get VL(0.039)=9.995 V rather than 9.8V
    With L=0.33H, I get VL(0.039)= 10.05 V

    My method ignores R1, as none of the current through R1 passes through L and vice versa.
    So I look at the series combination of L and R2 with a constant 12V applied across it from the battery.

    First I note that at t=0 no current is flowing through L and R2 and the full battery voltage appears across L.
    After a long time when the current through L is near enough constant, no voltage appears across L and the full 12V is across R2, giving us the maximum current in L and in R2 as 12V / 1.5 Ω = 8 A

    So the voltage across L starts at 12V then falls exponentially to zero and the current through L starts at 0 rising exponentially to 8A.

    The time constant for this is given as τ = L/R so for L and R2 in series, 0.32 H in series with 1.5Ω, τ=0.32/1.5 = 0.2133 seconds

    Therefore the voltage across the inductor, which starts at 12V at t=0, is given by VL(t)= VL(0) exp(-t/τ) or 12exp (-t/0.2133)
    So VL(0.039) = 12exp (-0.039/0.2133) = 12 exp(-0.1828125) = 12 x 0.83292 = 9.99509 V or 10 V to 2SF.

    So the voltage across R2 (not asked for) is 12 - 9,99509 = 2.00491 V or 2V to 2SF (since whichever Kirchoff law may apply, the battery maintains 12V across the LR series combination.)
    And then the current in both L and R2 being equal, it must be 2V/1.5Ω = 1.3A (calculator readout is 1.3366066)

    I can't see any of these values in your working, so i can't comment on any discrepancies.

    For part b when the switch is opened, any current from the battery ceases and all current now flows in the series circuit of R1, L, R2.
    There will obviously be a different time constant for the series circuit now, but the voltage and current will again vary exponentially with time.
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