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Physics 2 exam tomorrow. Need help with a couple more problems and im good!

  • Thread starter LakeMountD
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Physics 2 exam tomorrow. Need help with a couple more problems and im good! :)

1. a circuit has a 137 mH inductor across AB (the whole curcuit is a square with corners ABCD), a 50 ohm resistor across BC, and a 25 uF capacitor across CD with a 115V @ 60 hz power source between DA. What is the rms voltage across points AB, BC, CD, AC, and BD . I CANT FIGURE ANY OF THESE ONES OUT! They are driving me and like 4 other people crazy.


2. Two resistors are connected in series across a potential differencfe. Resistor A has twice teh resistance of resistor B. If the current carried by resistor A is I, then what is the current carried by resistor B? The answer is Just a) I but I can't seem to solve for it mathematically using relevant formulae.
 

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  • #2
dextercioby
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For the second problem,how is a series connection of resistors defined and what consequences does it have...?

For the first,what does "rms voltage" mean??

Daniel.
 
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dextercioby said:
For the second problem,how is a series connection of resistors defined and what consequences does it have...?

For the first,what does "rms voltage" mean??

Daniel.
rms voltage is root-mean square voltage.. I know in series the voltage changes but the current stays the same throughout the curcuit. I just do the proofs with the formulas to show it.
 
  • #4
dextercioby
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Conservation of electric charge insures the current stays the same in a series connection.

For the first problem,compute the circuit's impedance.Then the current (the amplitude) and then the tensions accross each side of the square.

Daniel.
 
  • #5
OlderDan
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LakeMountD said:
1. a circuit has a 137 mH inductor across AB (the whole curcuit is a square with corners ABCD), a 50 ohm resistor across BC, and a 25 uF capacitor across CD with a 115V @ 60 hz power source between DA. What is the rms voltage across points AB, BC, CD, AC, and BD . I CANT FIGURE ANY OF THESE ONES OUT! They are driving me and like 4 other people crazy.


2. Two resistors are connected in series across a potential differencfe. Resistor A has twice teh resistance of resistor B. If the current carried by resistor A is I, then what is the current carried by resistor B? The answer is Just a) I but I can't seem to solve for it mathematically using relevant formulae.
1) You have a series RLC circuit. You should have learned how to calculate the impedence of any single one of those three devices, and the impedence of the three of them in combination. The combination will allow you to determine the current being delivered by the power source, and the individual impedences will can then be used to find the voltages.

2) Charge is not created or destroyed, and ideal resistors cannot store charge. Where can the current flowing through RA go?

OOPs collision in cyberspace

We are both saying the same thing. Take your pick
 
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OlderDan said:
1) You have a series RLC circuit. You should have learned how to calculate the impedence of any single one of those three devices, and the impedence of the three of them in combination. The combination will allow you to determine the current being delivered by the power source, and the individual impedences will can then be used to find the voltages.

2) Charge is not created or destroyed, and ideal resistors cannot store charge. Where can the current flowing through RA go?

OOPs collision in cyberspace

We are both saying the same thing. Take your pick
on number 1- z=74 ohms
XL= 51.6 ohms

I= Vapp,peak/Z= 1.49 amps

Vrms,L = XL * I = 77 volts

and i did same thing for the resistor and capacitor and get right answer
HOWEVER- what do i have to do when going from AC or BD ? i tried combining the voltagse and then tried calculating a new voltage after first one and applying it to the second and both yielded wrong answers.

well on number 2 i understand why the answer is what it is. but proving it mathematically is another story. he wants it to come down to everything cancelling out in the end and the equations reduct to I1=I2 and we just cant seem to do the proof.
 
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  • #7
OlderDan
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I only have time for a quick comment here. What you must keep in mind is that the voltages accross the three devices in the circuit are not in phase. While at any point in time the true voltages in the loop must add up to zero, the rms voltages need not add up to zero. The current is the common thing to all elements in the circuit. That is why you must find the total current from the total impedence first, and then go back and figure out the individual voltages using that current.
 

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