Current in Electric Circuit

  • #1
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http://img475.imageshack.us/img475/2209/scan0010015za.png [Broken]

I'm supposed to find the current of the 8.0 ohm resistor.
I know that V=IR, P=IV, parallel circuits change in current and have constant voltage, and series circuits change in voltage and have constant current, but I get confused of parallel within parallel and whether to use the voltage/current at the point before or the EMF.
 
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Answers and Replies

  • #2
berkeman
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There are several ways to solve this but the most intuitive way for me would be to do the following:

-- simplify the whole resistor network down to one resistor to figure out the current supplied by the power supply. To simplify the resistor network, start at the innermost combination of resistors, and use the series and parallel combining rules to convert each pair of resistors into a single resistor. Continue until you end up with only the single equivalent resistor hooked around the 9V power supply.

-- Once you know the outside current, start again with the full circuit, and simplify the top 4 resistors into a single equivalent resistor. Now you can calculate what the voltage drop across the combination of the 4,6,8,10 Ohm resistors will be.

-- Starting again with the original circuit, and knowing what the voltage drop across the top combination of resistors is, how do you then figure out what the current is through just the 8 Ohm resistor? You'll still need to use some combination techniques to do it.
 
  • #3
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http://img528.imageshack.us/img528/5707/current23vc.jpg [Broken]

This is what I worked it out as - .333 A - that is probably wrong because I don't know if I'm supposed to used the total current or current in parallel to find the voltage. And this is awfully a lot of work for seems for a simple problem, is there an easier way? Seems like building the circuit yourself and using a multimeter would be easier.
 
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  • #4
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what I would do is convert the 4 and 8 ohm parallel resistors into an equivalent resistor, alpha. Then combine the 6ohm and the equivalent alpha resistor in series into another equivalent resistor, beta. Then combine the 10 ohm resistor and beta, which are in parallel into another equivalent resistor, gamma. Then you will have a .5 ohm, 5ohm, gamma omh, and 9 volt source all in series. Then calculate the current using KCL and KVL. Then you know the current at gamma, and work your way back in to whatever current you need.
 
  • #5
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i got it

plsease await 5 minutes i be back with the answer
 
  • #6
berkeman
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geoorge said:
plsease await 5 minutes i be back with the answer
Don't give him the answer -- this is a homework forum. You can give hints and tricks, but don't just give the answer. He won't learn to fish that way...., er, well you know what I mean.
 
  • #7
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In order to get the total current convert the circuit to an serial equivalent

1/(1/4o +1/8o) =2.66o ; here i break the first parallel

6o+2.66o = 8.66 ; here i merge two resitor in serial

1/(1/8.66o +1/10o) = 4.6o ; here i break the remaining parallel

now i got a serial circuit with 3 resitors

4.6o+5o+.50o = 10.14o and the Voltaje = 9volts

so the total I= .88 amp

so in a serial circuit all the current is the same so the voltaje must chance

so for the 4.6o(our resistor that we merge) the voltaje is

v=4.6 * .88 = 4.048 volts

now if we remember our original circuit has a resitor on the top with the value of 10o and the lower side has 8.6o (later we will return to the original form)

both have the voltaje of 4.048 volts but different current and in this point we break the .88 amps

I10o = 4.048/10o=.4048a
I8.6o=4.048/8.6o = .4674a

Now remeber that the 8.6 ohms were a resistence of 6 +2.66(first parallel)

V6o = .4674a * 6ohms = 2.8044volts
V2.66= .4674*2.6ohms =1.2152 volts (v6o+v2.66o = 4.048 volts)

V10ohms = .4048a *10 ohms = 4.048 volts

I10o+I(6o&2.66o) = .4674+.4048 = .88a

now we have that the volt that gets to the 2.66o is 1.121 volts


so

I8o= 1.2152/8 = .1519
I4o= 1.2152/4 = .3038

:)
 
  • #8
berkeman
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I guess the fish got away....
 
  • #9
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click edit--->delete---> delete message -->delete this message

its not too late.
 
  • #10
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berkeman

You are right, sorry i will try to let students to find the answers, in this case i just, show the answers so he can easly track the problem, in think that in circuits the most important thing is not the formula but the drawing
 
  • #11
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Thanks for the help george, I get it now. I was thinking of the problem as one big picture and not taking it apart piece by piece. Also I didn't think the IR on a resistor was not only the voltage drop after it but also the "flow" I guess through it.
 

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