Register to reply 
Load resistor in circuit: Find the noload voltage. 
Share this thread: 
#1
Jan1813, 04:20 PM

P: 301

1. The problem statement, all variables and given/known data
A. Find the noload voltage of V_{0} 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? 2. Relevant 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)] 3. The attempt at a solution I forgot how to deal with loadresistors, 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) = 72V Still need help for other parts though 


#2
Jan1813, 04:55 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,765

The load resistor is treated exactly the same as any other resistor.
I cannot help with the circuit without seeing it though. 


#3
Jan1813, 06:12 PM

Mentor
P: 40,944

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


#4
Jan1813, 07:06 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,765

Load resistor in circuit: Find the noload voltage.
Ah  I can see it now.
It didn't show up when I posted. The questions AD 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
Jan1913, 06:27 PM

P: 301

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 = 0.48W Assuming 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
Jan1913, 11:27 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,765

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
Jan1913, 11:51 PM

P: 301

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
Jan2013, 06:58 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,765

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
Jan2113, 06:37 AM

P: 301

And there's no current through shortcircuited resistors either then? and I get this error when trying to download the circuitikz files: Thanks again anyways though. 


#10
Jan2113, 07:05 PM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,765

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/w...vscircuitikz/ ... review with links. 


Register to reply 
Related Discussions  
Thevenin circuit Max power across Load resistor  Introductory Physics Homework  3  
Voltage across a load resistor in a Voltage Divider Circuit  Introductory Physics Homework  6  
Load voltage with shorted load / open load  Introductory Physics Homework  6  
Linear Circuit  current through Load Resistor  Engineering, Comp Sci, & Technology Homework  8  
Maximizing voltage across a load resistor?  Introductory Physics Homework  4 