# RL DC Circuit

1. Mar 22, 2015

### throneoo

1. The problem statement, all variables and given/known data
(see figure)

The switch is initially open for a long time.
At t=0 it is closed, and after 1ms it is reopened.
Find the voltage across Resistor 2 at (i)0<t<1ms and (ii) t>1ms.

2. Relevant equations
V0=120V
R1=500 Ω
R2=125 Ω
Inductance of the inductor = 0.2 H

3. The attempt at a solution
(i)

Taking reference points across the inductor, the Thevenin circuit has applied voltage 24V and resistance 100 Ω.
The inductor tries to oppose the sudden increase in current and generates an emf with voltage

24*exp(-RThevenint/L)24*exp(-100t/0.2)=24*exp(-500t).

Back to the original circuit, by Kirchoffs voltage law, the voltage across Resistor 2 = that of the inductor =24*exp(-500t) V
with an anti-clockwise current
(ii)

At t= 1ms, the voltage across Resistor 2 = 24*exp(-0.5) and the current is (24/125)*exp(-0.5) anticlockwise....

I stopped here due to the following questions.

Question:
(i)
Does the applied voltage contribute to the current flowing through Resistor 2 at all? Maybe it does only at the moment the switch is closed, when the inductor behaves like an open circuit by generating 24V to oppose the applied voltage. After this instant, the inductor has no resistance at all and it shorts out Resistor 2.

How can I apply superposition theorem in this problem? How do I isolate the effect of the inductor? Something tells me it's not as simple as replacing it as an open/close circuit because of its time dependent behavior. This concerns how I find the current in each circuit element as a function of time.

(ii)

When the switch is reopened, does the transient response from the closing of the switch still remain?

If that is the case , there would be a superposition of response across the inductor/resistor.
Otherwise, the new response would still have to counteract the current from Resistor 2 and the absence of the voltage source.

But how do I find the new response?

The inductor would produce a voltage to maintain the initial current across it, which is 0.24*(1-exp(-0.5)) (downward/clockwise) but what's next?

#### Attached Files:

• ###### m.png
File size:
669 bytes
Views:
100
2. Mar 22, 2015

### Staff: Mentor

How do you figure that the current through R2 is anticlockwise? Around what loop?
I agree with the voltage across R2 at t = 1ms. I'm not sure that the current through R2 will be particularly helpful; You'd be better off knowing the current through the inductor. I still don't know what "anticlockwise" refers to.
No. The inductor gradually allows more current to flow through it and the back-emf across the inductor will gradually fall from its initial 24 V value, heading towards zero via exponential decay.

So the voltage across the inductor decays while the current through the inductor increases, both according to exponential curves specified by the circuit time constant.
Hint: Find the current flowing through the inductor at the instant the switch opens. That current will continue to flow through the inductor (and in the same direction) in the instant after the switch has opened...
Yes. Whatever current was flowing in the inductor at that instant remains.
Terminology nit-pic: Current goes through, not across.

Draw the new circuit after the switch opens. Where will that initial inductor current flow?

3. Mar 22, 2015

### throneoo

around the inductor(right loop),since i thought the back emf is the only voltage source to resistor 2. but now that i think of it there is no reason this current does not go back to the battery in a clockwise direction around the left loop
according to the thevenin circuit the currect through the inductor at 1ms is 0.24exp(-0.5)

i meant the exponential decay part. but i think this already gives me the answer

the new circuit only consistS of the inductor and resistor 2 in a single loop. if the only current that remains is the one through the inductor it will go clockwise with magnitude
0.24exp(-0.5) A. voltage across the resistor is the product of this current and its own resistance

4. Mar 22, 2015

### throneoo

edit: the voltage across the resistor should be 0.24exp(-0.5)*R2*exp(-R2t/L)

5. Mar 22, 2015

### Staff: Mentor

How do you figure that? Does the current through the inductor increase or decrease with time?
The circuit description is correct, but you'll want to check your value for the initial current.

The voltage across R2 the instant after the switch opens will depend upon the current that flows through it. Note that it can jump radically from the value it had the instant before the switch changed! What current flows through the resistor in the instant after the switch opens? What direction?

And also note that the time variable can't be the same as the previous time variable, as the "new" t = 0 occurs 1 ms after the original t = 0. So either use a new variable for t here or bias it with the appropriate offset.

6. Mar 23, 2015

### throneoo

When the switch is closed, the thevenin circuit consists of an applied voltage of 24V the inductor and a resistor of 100 Ω, all connected in series.

i realized i missed something.
the current is VTh/RTh *(1-exp(-500t),which increases with time. therefore at 1ms, it should be 0.24*(1-
exp(-0.5))
same as the above
same as the above,downward through the inductor.
forgot to edit that part. the t should be replaced by t-1ms