How Do You Calculate Instantaneous Power in a Resistor with Switching Circuits?

[Ronaldo95163]

Homework Statement

Given the following switch positions in time, sketch a graph of the instantaneous power dissipated in the 50 ohm resistor with time.

Homework Equations

P=I^2R=V^2/R=IV

The Attempt at a Solution

Basically I treated the inductor as a short circuit and obtained the current through the circuit as 0.4A so the power dissipated in the 50ohm resistor would be I^2R = 8W

For the Capacitive circuit I used KCL to get a differential eqn:
20-v/50 = v/1x10^-3 + (1x10^-3)dv/dt
Solving that I got v = 400/21e^-21t+400/21

which I then squared and divided by R = 50 to obtain an expression for power which I used to sketch the graph.

 

[Svein]

Basically I treated the inductor as a short circuit and obtained the current through the circuit as 0.4A so the power dissipated in the 50ohm resistor would be I^2R = 8W

That is a bit simplistic. The rule of thumb for an inductor is that it "resists changes in current". Thus, the inductor will have an effect whenever the switch changes state.

 

[NascentOxygen]

For the capacitor voltage, your badly-named constant, C, should be a negative value.

 

[Ronaldo95163]

Sorry about that wrt the constant...I just used a calculator to get the gen soln to save time. But yes it should be.
@Svein since the circuit is initially with the switch closed...what would be the it's effect?
What a tried was a thevenin equivalent looking into the terminals of the inductor and I got a 25\3 ohm resistor in series with the inductor and a 10/3 volt source.

Writing a DE for it and solving it I got the answer shown and using t=0 i=0.4 it resulted in that particular integral.(Edit c should be 0 in the case...therefore i would simply be equal to 0.4)

So the power eqn formed would i^2r which still led me to getting 8W

 

[gneill]

My suggestion would be to draw the two configurations for each of the subcircuits (connected to the source via the switch and isolated from the source) and determine the time constants that pertain to each. That's a total of four time constants to determine. Replacing the voltage source and resistor networks with Thevenin equivalents will be advantageous for determining the time constants for those cases.

Take a close look at all the time constants and see which ones, if any, are going to be problematical given the switch timings. Remember the "##5 \tau## rule" for time constants.

Knowing the time constants and initial and steady state end states should allow you to sketch the results fairly easily.

 

[Ronaldo95163]

Thanks man ill give it a shot. Just got home from school. The equation I got for the current through the inductor...would that be the same as that through the 50ohm? We were told in class that they act as short circuits so I am assuming that the equation i got there from the thevenin circuit would represent the current through the 50 ohm resistor as well

 

[gneill]

Thanks man ill give it a shot. Just got home from school. The equation I got for the current through the inductor...would that be the same as that through the 50ohm? We were told in class that they act as short circuits so I am assuming that the equation i got there from the thevenin circuit would represent the current through the 50 ohm resistor as well

The current through an inductor will change over time. An inductor will look like an open circuit to sudden changes, and a short circuit eventually when the circuit settles into steady state.

Compare that with a capacitor which looks like a short circuit to sudden changes and an open circuit at steady state.

Check the defining equations for these components.

 

[Ronaldo95163]

Thanks man.
I worked it over and this was my solution:

 

[NascentOxygen]

Inductor current is initially 0.4A. Once source is disconnected, the inductor current falls exponentially with τ of 1/100 sec. So no negative sign in front of the 0.4

Is it known whether the horiz axis of your switching sequence graph is in seconds or 1/100 sec? (Maybe this is a trick question, and there is no need to involve the exponentials?)

 

[Ronaldo95163]

Sorry about that.
No it wouldn't because it is disconnected from the rest of the circuit containing the 50ohm resistor