Load resistor in circuit: Find the no-load voltage.

In summary: Many people have heard the term "short circuit" but don't realize what is involved."short circuit" is a wire, "open circuit" is a gap... usually means "no load".So, in part C, the circuit only has one resistor... well done.
  • #1
Color_of_Cyan
386
0

Homework Statement



http://imageshack.us/a/img62/5467/homeworkprob14.jpg

A. Find the no-load voltage of V0

B. Find Vo when RL is 450 kΩ

C. How much power is dissipated in the 30 kΩ resistor if the load terminals are accidentally short circuited

D. What is the maximum power dissipated in the 50k resistor?

Homework Equations


V = IR

Voltage Division:
(Voltage across series resistor) = [(resistance) / total series resistance)](total input V)

Current Division (for 2 parallel resistors):
(current across parallel resistor) = [(other resistor) / (sum of parallel resistors)](total incoming current)]

The Attempt at a Solution



I forgot how to deal with load-resistors, and not sure how they apply to a circuit like this.

I don't need help with part B because, well, that's straight forward, and you find parallel resistance first between the load and other one by V

because: (1/450 + 1/50)^-1 = 45Ω

then using voltage division when you have only 30k and 45k in series:

V knot = [45Ω / (45 + 30)Ω] (120V) = 72VStill need help for other parts though
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
The load resistor is treated exactly the same as any other resistor.
I cannot help with the circuit without seeing it though.
 
  • #3
Simon Bridge said:
The load resistor is treated exactly the same as any other resistor.
I cannot help with the circuit without seeing it though.

I'm able to see the circuit, Simon. Is ImageShack blocked by your firewall? Here is the URL if it helps:

http://imageshack.us/a/img62/5467/homeworkprob14.jpg

.
 
Last edited by a moderator:
  • #4
Ah - I can see it now.
It didn't show up when I posted.

The questions A-D can be understood by redrawing the circuit diagram each time according to the description. i.e. for A, just erase the load resistor part. for C - how does "accidentally short circuiting the load terminals" change the circuit diagram?

I think there is an equation missing too - relating power to voltage and resistance.
 
  • #5
Yeah I kept editing my post at first to change it to correct info because I got this mixed up with another problem (although similar). Sorry about that. So anyways...
That would be P = V^2/R added to the equations then.For part C, I set the circuit to be the same as in part A (with just 30k and 50k in series) and then did voltage division for the 30k resistor:

V (30k) = (30k / 30k+50k)(120V) = 45V

then

P = V^2 / R = (45)^2 / 30000 which makes it 0.1125 W but this isn't the same as the given correct answer.

The correct answer would mean that V = 120V when doing P = V^2 / R which is (120)^2 / 30000 = 0.48WAssuming the given correct answer for part C really is correct, then why do you have to use 120V for P = V^2 / R? I got part D though, just P = 75^2/50000 and then P = 0.1125 W. 75 V because that's the potential across the 50k resistor found earlier in part A.
 
  • #6
For part C, I set the circuit to be the same as in part A (with just 30k and 50k in series) and then did voltage division for the 30k resistor

That's what you get if the load terminals are open circuit.
Which is part A.

The question wants:

C. How much power is dissipated in the 30 kΩ resistor if the load terminals are accidentally short circuited

... a short circuit means you put a wire across the load resistance.
 
  • #7
Alright, I admit I didn't know the difference (now I'm sad, heh). I was never really told about that (enough) in my physics class. I need to brush up on this then too, so if you have another wire that bypasses a resistor that means you can disregard potential drop across that resistor then, right (because here the wire across the load resistor would then also bypass the 50k ohm resistor too, right)? Assuming it's the path of least resistance, though. Thanks for all your help, by the way.
 
  • #8
Many people have heard the term "short circuit" but don't realize what is involved.
"short circuit" is a wire, "open circuit" is a gap... usually means "no load".

So, in part C, the circuit only has one resistor... well done.

Aside:
You notice that you can draw a better circuit diagram freehand than with a computer?
Price of progress I guess - LaTeX comes to the rescue there too:
http://www.texample.net/tikz/examples/circuitikz/
... don't think PFLatex supports it though.
 
  • #9
Simon Bridge said:
Many people have heard the term "short circuit" but don't realize what is involved.
"short circuit" is a wire, "open circuit" is a gap... usually means "no load".

So, in part C, the circuit only has one resistor... well done.

So that would be a yes then (in the sense that you really can disregard potential drop across a resistor anytime if the current can go around the resistor in another path (short circuited))

And there's no current through short-circuited resistors either then?
Simon Bridge said:
Aside:
You notice that you can draw a better circuit diagram freehand than with a computer?
Price of progress I guess - LaTeX comes to the rescue there too:
http://www.texample.net/tikz/examples/circuitikz/
... don't think PFLatex supports it though.
and I get this error when trying to download the circuitikz files:

Not Found

The requested URL /mredaelli/about.html was not found on this server.
Thanks again anyways though.
 
  • #10
And there's no current through short-circuited resistors either then?
Sometime the term "sort circuited" is used incorrectly to refer to a resistor (or other component) that has blown up (or otherwise fatally damaged) as a result of a short circuit - somewhere.

A blown up resistor is likely to be an open circuit.
Either way - no current.

Rats about the circuitiks files ... you are right, link seems to be out of date.
I get mine with gnu/linux and Mac users get it with MacTex by default, I keep forgetting how annoying Windows can be.

There is a ctan page about them.

Checking:
http://web.ics.purdue.edu/~cdelker/wp/2012/08/circuit_macros-vs-circuitikz/
... review with links.
 

Related to Load resistor in circuit: Find the no-load voltage.

1. What is a load resistor?

A load resistor is an electronic component that is used to add resistance to a circuit, often to limit the current flow or to dissipate excess power.

2. How does a load resistor affect the circuit?

A load resistor affects the circuit by reducing the amount of current that flows through it. This can help regulate the voltage and prevent damage to other components in the circuit.

3. Why is it important to find the no-load voltage in a circuit?

It is important to find the no-load voltage in a circuit because it can help determine the overall health and functionality of the circuit. The no-load voltage is the voltage measured across a circuit when no current is flowing through it.

4. How do you calculate the no-load voltage?

The no-load voltage can be calculated by using Ohm's law, which states that voltage is equal to current multiplied by resistance (V=IR). You would need to know the resistance of the load resistor and the current flowing through the circuit to calculate the no-load voltage.

5. What factors can affect the no-load voltage in a circuit?

The no-load voltage in a circuit can be affected by several factors, including the resistance of the load resistor, the type and quality of other components in the circuit, and the voltage source supplying the circuit. Temperature, humidity, and other environmental factors can also impact the no-load voltage.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
17
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
10
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
17
Views
10K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
0
Views
845
Back
Top