Resistance with resistors between parallel resistors

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
TiernanW
Messages
25
Reaction score
2

Homework Statement


The overall resistance of the circuit is 12.5Ω. Calculate the value of the resistor labelled R in the circuit.
14iertg.png


Homework Equations


1/R = 1/R1 + 1/R2
R = V/I

The Attempt at a Solution


I understand that you use the first equation for parallel resistors and the second to calculate resistance from voltage and current. The bit I am having trouble with is the network of 4 resistors on the left hand side. The resistance on the right hand-side is 1/R = 1/2 + 1/6, so R = 1.5. So then 12.5 = 6 + 1.5 + the resistance of the left hand-side network.

I just need guidance on how to calculate it because it puzzles me as there is 2 resistors between the other 2 in parallel.
 
Last edited:
on Phys.org
TiernanW said:
2 resistors between the other 2 in parallel.
Are you sure?
Hint:Current through series resistors is same.
 
I find it helpful to follow the current around the circuit. Start at the battery. As the current flows where does it split? It's enough to start at the point in the circuit where you lost the bubble.
 
Thanks guys! Just considered the other 3 (5, 1 and R) as a series and the 10 as parallel with those. :)
 
  • Like
Likes   Reactions: CWatters
TiernanW said:
The resistance on the right hand-side is 1/R = 1/2 + 1/6, so R = 1.5. So then 12.5 = 6 + 1.5 + the resistance of the left hand-side network.

Yes that's ok.

TiernanW said:
Thanks guys! Just considered the other 3 (5, 1 and R) as a series and the 10 as parallel with those. :)

Yes that's ok for the "left hand side network".

Just turn it all into an equation and solve for R