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Voltage drop and current for each resistor 
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#1
May810, 09:31 PM

P: 8

1. The problem statement, all variables and given/known data
heres a picture of the problem. Please dont give me the answer just tell me what are the steps to complete it. thanks. http://i307.photobucket.com/albums/n...download1.jpg 2. Relevant equations V=IR 3. The attempt at a solution 1/6+1/3=2 Ohms 


#2
May810, 10:34 PM

P: 62

Ok, so you now have two resistors in the circuit you can calculate the voltage drop across each resistor and, using Ohm's law, the current. You need to be a little more specific with what you're looking for so you can get more help.



#3
May810, 11:09 PM

P: 8

Im looking to find those items. The voltage drop and the current. I dont know how.



#4
May810, 11:52 PM

P: 41

Voltage drop and current for each resistor
Since there is one indepedent voltage source of 20 Volts that means the voltage drop across all the resistors should equal 20V.
I would have made this look nice but the LaTex Isn't working. First Combine the Resistors in Parrallel: 1/R =(1/R_1)+(1/R_2)+...(1/R_N) 1/R = (1/6 Ohm) +(1/3 Ohm) 1/R = (1/6 Ohm) + (2/6 Ohm) 1/R = (3/6 Ohm) = (1/2 Ohm) 1/(1/2 Ohm) = R = 2 Ohm So now you have ^^^^^^^^ ___________________ Sorry for the bad drawing (  = voltage source, ^^^^ = resistor) Now combine the resistors in series R= R_1 +R_2+...R_N R= 3 Ohm + 2 Ohm R = 5 Ohm So now you have ^^^^ ____________ Using V = IR you can solve for the current: V=IR (20 Volts) = I * (5 Ohm) I = 4 Amperes Now split the circuit back up so that it is 2 resistor in series. Resistors in series have the same current ^^^^^^^^ ___________________ So for the first resistor (3 Ohm resistor) V = IR I = 4 Ampere R = 3 Ohms V = (4 Ampere)(3 Ohm) V= 12 Volts Voltage on the 3 Ohm resistor directly right of the independent voltage source is 12 volts Now for the second resistor (2 Ohm Resistor) V = IR I = 4 Ampere R = 2 Ohms V = (4 Ampere)(2 Ohms) V = 8 Volts Voltage on the 2 Ohm resistor directly right of the 3 Ohm resistor is 8 volts. Split it up so you are back with the circuit you were first with. ^^^^.^^^^ ___________^^^^ Resistors in parrallel have the same voltage so: 6 Ohm Resistor in Parrallel: V=IR (8 volts) = I * (6 Ohm) I = (4/3) Amperes 3 Ohm Resistor In parrallel: V= IR (8 Volts) = I * ( 3 Ohm) I = (8/3) Ampere So, Resistor 1: Resistor(R)= 3 Ohm Voltage Drop(V) = 12 Volts Current(i) = 4 Amperes Resistor 2: Resistor(R) = 6 Ohm Voltage Drop(V) = 8 Volts Current(i) = (4/3) Ampere Resistor 3: Resistor(R) = 3 Ohm Voltage Drop(V) = 8 Volts Current(i) = (8/3) Ampere 


#5
May910, 10:27 AM

P: 62

Err, fallen186, you may want to read the rules over real quick. Specifically this line:



#6
May910, 11:53 AM

P: 22

Anyway back to fallen186īs answer. Thank you. That helped me as well. 


#7
May910, 11:54 AM

P: 62




#8
May910, 11:57 AM

P: 22

Oh well every forum has its own rule. But still its not like he is posting his whole paper or HW here, he is only posting one question so wouldnt that be something else? I mean mostly you can only find how something works by knowing the answer and the exact way of getting there. Hence you need the full explanation? 


#9
May910, 02:03 PM

P: 8

Yeah he was very helpful to me too. I deff learned how to do the problem and thats what i wanted. Thanks man!



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