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I am working on a problem involving a resistor and inductor in parallel with eachother - I am trying to work out the voltage across the inductor as a function of time. It is clear to me that I will need a formula for 0 < t < 1ms and one for t > 1ms (Because the switch is closed then opened again) however I am really struggling to get my head around how I will work this out. If someone would be kind enough to point out the logical steps for this problem it would be much appreciated - I am wondering if I need to work out a thevenins equivalent circuit then try to solve, however when I attempt this I find the resistance would change with time which wasnt helpful in solving. I have tried my textbooks and google but I couldn't find a worked solution or notes on how to solve such a circuit. Many thanks!

Circuit diagram

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  • \$\begingroup\$ You might try Thevenin's for the first 1mS, put a current source in parallel with 500ohms, 125ohms and L. Then solve separately for what happens after 1mS, with the current in the inductor continuous at 1mS. \$\endgroup\$ – Neil_UK Mar 19 '16 at 16:48
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    \$\begingroup\$ The keyword is: "transient state". Don't use Thevenin, to solve this circuit use classic Kirchoff's laws, but remember that voltage across the inductor is a function of inductance, current and time U=L*(di/dt). Another method (IMHO much simpler) is to use Laplace transform. \$\endgroup\$ – Jakub Rakus Mar 19 '16 at 17:37
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A key thing to remember is that the current in an (idealised) inductor cannot change instantaniously but the voltage can (this is the opposite of the ideal capacitor where the voltage cannot change instantanously but the current can)

There are three stages to consider (I will assume all variables use si units).

Before t=0:

The voltage and current in the inductor are zero.

Between t=0 and t=10-3

We can replace the voltage source and two resistors with a thevin equivilent circuit. This works out to a 24V supply with a 100 ohm series resistance. We now have a series LR circuit supplied by a 24V supply with zero current at t=0

We could solve the series LR circuit from first principles but I just went and looked it up on google and found http://www.electronics-tutorials.ws/inductor/lr-circuits.html . This tells me that the equation for this case is.

$$I=\frac{V}{R}(1-e^{-Rt/L})$$

Where I is the current in the indctor, L is the inductance and V and R are the 24V and 100 ohm we got from our thevinin calculation above.

After t=10-3

The only components relavent now are the inductor and the 125 ohm resistor. The current in the inductor at the start of this phase will be the same as the current in the inductor at the end of the previous phase but it is now forced to go through the resistor, this will cause the voltage on the inductor to reverse and the energy stored in the inductor to discharge. I'll leave you to either work out and solve the differential equation or google for the equation needed in this case.

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For t < 0:

Since it was left open for "a long time", then its safe to assume that the inductor has been completely drained of energy. So \$V_L = 0\$.

For 0 < t < 1ms:

Taking the Laplace transform of the components and simplifying (as a voltage divider):

$$V_L(s) = \dfrac{120sLR_2}{R_1R_2 + sLR_1 + sLR_2}$$ $$= \dfrac{24s}{s + 500} = 24\dfrac{s}{s + 500} = 24(1 - \dfrac{500}{s + 500})$$

Inverse Laplace gives:

$$V_L(t) = 24(\delta(t) - 500e^{-500t})$$

Since t > 0, this results in:

$$V_L(t) = - 12000e^{-500t}$$

For t > 1ms:

The inductor drains all its stored energy through the resistor using the standard equation.

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