Voltage drop and current for each resistor

AI Thread Summary
The discussion focuses on calculating voltage drop and current for resistors in a circuit. Participants outline the steps to combine resistors in parallel and series, applying Ohm's Law (V=IR) to find the total resistance and current. The voltage drop across each resistor is calculated, revealing specific values for each component in the circuit. There is also a mention of forum etiquette regarding providing complete solutions to homework questions. Overall, the conversation emphasizes understanding the process rather than just obtaining answers.
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Homework Statement


heres a picture of the problem. Please don't give me the answer just tell me what are the steps to complete it. thanks.

http://i307.photobucket.com/albums/nn296/alan1592/download-1.jpg

Homework Equations



V=IR

The Attempt at a Solution



1/6+1/3=2 Ohms
 
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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.
 
Im looking to find those items. The voltage drop and the current. I don't know how.
 
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
 
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Last edited by a moderator:
Cheesus128 said:
Whats the point of that rule?
So the members here don't do your homework
 
Fronzbot said:
So the members here don't do your homework

Hahahaha that made me smile
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 wouldn't 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?
 
Yeah he was very helpful to me too. I deff learned how to do the problem and that's what i wanted. Thanks man!
 
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