RL, DC circuit with a resistor in parallel

Question

I've been revising for an upcoming exam and keep coming across this format of question, A coil with internal resistance N, is set in parallel with a resistor M, both in parallel with a voltage source and switch.

• a) Determine the Coil current at time S, after the switch is closed.

• b) After some time the current has reached its maximum value, what is the magnitude of the inductor's self-induced EMF at the moment of re-opening the switch?

^{simulate this circuit – Schematic created using CircuitLab}

At 20ms after the switch is closed, the Coil's current is 0.8647A; However, I don't know how we've ended with this result.

Time Constant
I've assumed that the time constant is seen from the voltage source, meaning R1/R2 are in parallel, giving a Resistance total of 50 ohms, a time constant of 1/50, or 20 ms.
However, I've recently seen that I'm meant to consider the voltage source “shorted” and the inductor to be where I view the circuit from, meaning that R1 is completely missed, resulting in a 1/100 or 10ms time constant.

Equation
The equation I'm using is \$i = I(1-e^-(t/\tau)) \$,though this requires me to have an initial current of 1, my understanding is that I find the current by determining the resistors in parallel of the circuit, ignoring the coil's inductance. However, that would leave me with a current of 2 Amps. The only alternative is that I again ignore the parallel resistor and consider only the coil's resistance?

My confusion is at the method that is needed to approach this from in terms of R1/R2, as they're both the same here, but if they were different values, I have no idea what I'm doing.

Additionally, at the moment of the switch opening, the magnitude of the inductor's self-induced EMF is 200 V. I have no idea why, though. Is it because R1 and R2 go from parallel to series? Both resistors have 100 V in parallel, but now they're in series \$ V = V_r + V_l\$?

I'm really unfamiliar with this configuration, and unsure where to look for further information. The past papers I have access to have only some numerical answers, but no explanation of the method.

Any help would be appreciated!

Answer 1

When the switch opens, inductor current can't change instantly. In other words, immediately after the instant of switch opening, inductor current is the same as it was immediately prior to the switch opening. That's what inductors do.

This can be modelled in two stages, the DC state with the switch closed, after all currents and voltages have settled (below left), and the state with the same current flowing but with the switch open (right):

^{simulate this circuit – Schematic created using CircuitLab}

I've changed R1 to 200Ω because with 100Ω the voltage across R1 doesn't change in magnitude when the switch is opened, and the effect is not well illustrated. I invite you to check this for yourself.

R2 and L1 represent a single inductor with coil resistance 100Ω.

The left circuit represents the state some time after the switch is closed. The voltage applied across L1+R2 is 100V, and inductor current will rise to and eventually settle at its maximum:

$$ I_0 = \frac{V_1}{R_2}= \frac{100V}{100\Omega} = 1A $$

This will be the final current with switch closed, but following switch closure there will be an exponential rise of current from zero to 1A, with time constant \$\tau = \frac{L_1}{R_2} = \frac{1H}{100\Omega}=10ms\$:

$$ I = I_0 \left( 1 - e^{\frac{-t}{10ms}} \right) $$

Setting \$t = 20ms\$:

$$ \begin{aligned} I &= 1A \left( 1 - e^{\frac{-20ms}{10ms}} \right) \\ \\ &= 0.865A \\ \\ \end{aligned} $$

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When the switch is opened (at time \$t=0\$), the inductor prevents an instantaneous change of current, becoming itself a current source of 1A (which will of course decay over time). The switch is open, meaning that the only path for that current is upwards via R1. This is the situation depicted above right.

We must be consistent with the sign of current in R1, which indicates direction. Initially current through R1 would have been downwards (positive), but this new post-switch-opening current is upwards, and will therefore have a negative sign:

$$ V = -I_0R_1 = -1A \times 200\Omega = -200V $$

That's shown on voltmeter VM1. Immediately following switch opening, there will be 200V across R1, with the lower potential at the top.

Following the switch opening, current will decay from 1A to zero at a rate determined by the resistance around the loop, which is now \$R_1 + R_2 = 300\Omega\$. The time constant will be \$\tau = \frac{L_1}{R_1 + R_2}\$, and current will decay according to

$$ I = I_0 e^{\frac{-t}{\tau}} $$

where \$I_0 = 1A\$. You are interested in voltage across the combination of L1 & R2, which is the same as the voltage across R1, found with Ohm's law:

$$ \begin{aligned} V_{R1} &= -IR_1 \\ \\ &= -R_1I_0 e^{\frac{-t}{\tau}} \\ \\ &= 200 \times -1 \times e^{\frac{-t}{300}} \\ \\ &= -200 \times e^{\frac{-t}{300}} \\ \\ \end{aligned} $$

Answer 2

This analysis is only valid for those who know the theory and use of the Laplace transform in the study of linear electric circuits. In non-electrical courses, probably, this study is optional. In any case, let me publish the results:

Answer 3

When switch is closed the coil is charged from V1 through just series R2. The R1 just draws a constant 1A from V1 and does not affect the coil charging process.

When switch becomes open you can remove V1 from equation and coil is discharged through series R1+R2. So if the coil was fully charged (to 1A) during switch closed a right after switch becomes open the voltages across R1 and R2 are follows:

Then the voltages Vr1 and VR2 continously decreses to zero volts.