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Finding the current through the resistor! i've tried so many different ways!got some.

  • Thread starter mr_coffee
  • Start date
  • #1
1,629
1
Hello everyone, i'm confused on what i'm suppose to do here...the problem says:
R1 = 100 . R2 = R3 = 45.0 . R4 = 75.0 , and the ideal battery has emf = 6.00 V.
Here is the image:
http://www.webassign.net/hrw/27-42.gif
Here is the question:
(a) What is the equivalent resistance?
117.31
(b) What is i in R1?
.051
(c) What is i in R2?

(d) What is i in R3?
A
(e) What is i in R4?
A

Okay i got (a) and (b), but messed up (c) like alot of times...
I tred i = E/R1+R2, i = .051, and a few others, all wrong...any ideas what i'm doing wrong?
 

Answers and Replies

  • #2
382
2
r1 is not in series with r2 so your method can't work. r2, r3 and r4 are in parallel, find the equivalent resistance, r_eq1.
now, r_eq1 is in series with r1, so you can find the voltage through r_eq1 which is the same as r2,r3 and r4 since they are parallel.
have the value of r and the voltage, then use Ohm's Law to find the individual current.
 
  • #3
1,629
1
Hey thanks for the reply!! So your saying, find the rquivalent resistance of r2,r3, and r4? I did that and got 225/13, I knew R1, so i added r1+225/13 = 117.31, and i found the current through R1, = .051, you talk about finding the voltage through R1_eq1, R = V/i, so i = V/R, i = 6/R2? i = 6/45 = .23333 amps for the current through R2, like that?
 
  • #4
382
2
sorry, actually you don;t need to find r_eq1 since you know i1=0.51 A already.
the voltage across r_eq1 = 6-i1*r1.
this voltage is the same for r2,r3 and r4 which are parallel.
you know their resistances and the voltage, use ohm's law afterwards.
 
  • #5
1,629
1
thank you! But i get a negative voltage r_eq1 = 6-.51*100 = -54, is that okay?
 
  • #6
382
2
i didn't check the anwers for the first 2 parts.
r_eq of the circuit = r1+255/13 = 100 +255/13=119.6 ohm.
i_r1= v_battery/r_eq = 6/119.6 = 0.0502 A. : after finding the r_eq1 for r2,r3,r4, you now get a series circuit consists of r1 and r_eq1.
 

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