Engineering Current Divider Circuit (Find a Resistance)

AI Thread Summary
The discussion focuses on solving for the resistance R in a current divider circuit where resistors are primarily in parallel, not series. The user initially calculated R as 720 ohms, while the textbook states it should be 30 ohms, leading to confusion. Clarifications indicate that the 60-ohm resistor is not in series with the 40-ohm and 80-ohm resistors, and the total current of 20 A must be divided between the parallel branches. The importance of correctly applying the current division formula is emphasized, particularly ensuring the numerator is accurate. The user is encouraged to re-evaluate their calculations based on these insights.
johnsmith7565
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Homework Statement
Find The Value Of R That Will Carry a 4 A Of Current To Flow Through The 80 Ohm Resistor In The Circuit Shown.
Relevant Equations
Parallel Circuit rules and Series circuit rules, i0 = (R1)/(R1+R2)is, where “is” is the source current, R1 and R2 are the resistances, and i0 is the current across a resistor.
The resistors are all in series , with exception of R, so I added them together and then used the current divider equation to solve for R, and I got 720 ohms. The textbook says R should be 30 ohms. I’m completely lost.
 

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Hi @johnsmith7565 ,

Welcome to Physics Forums.

Note that the ##60 Ω## resistor is not in series with the ##40 Ω## and ##80 Ω## resistors; The existence of the resistor ##R## precludes that.

What you can say is that 20 A enters the top node from the direction of the source, and that it must divide between the two parallel branches. Write your current division expression so that it divides the 20 A as required using only the three resistors of the parallel branches.
 
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gneill said:
Hi @johnsmith7565 ,

Welcome to Physics Forums.

Note that the ##60 \Ohm## resistor is not in series with the ##40 \Ohm## and ##80 \Ohm## resistors; The existence of the resistor ##R## precludes that.

What you can say is that 20 A enters the top node from the direction of the source, and that it must divide between the two parallel branches. Write your current division expression so that it divides the 20 A as required using only the three resistors of the parallel branches.
I still don’t understand. So the 40 ohm resistor and the 80 ohm one are in series so I add them, and now my equation is 4=(20)(120/R+120) and R is still too large.
 
johnsmith7565 said:
I still don’t understand. So the 40 ohm resistor and the 80 ohm one are in series so I add them, and now my equation is 4=(20)(120/R+120) and R is still too large.
I think you'll want to take a closer look at the current divider expression. Let's say you have two parallel resistances A and B with some total current ##I## being split between them. Then you can write:

##I_A = I \cdot \frac{B}{A + B}##

##I_B = I \cdot \frac{A}{A + B}##

Make sure that you get your numerator correct!
 
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