Solve the Circuit: Find R1-R7 to Calculate A to B Resistance

  • Thread starter Thread starter spoonthrower
  • Start date Start date
  • Tags Tags
    Circuit Resistance
AI Thread Summary
The discussion focuses on calculating the equivalent resistance between points A and B in a given circuit with resistances R1 to R7. The user correctly identifies that R4, R5, and R6 are in parallel, calculating their combined resistance as 2.72 ohms. However, confusion arises regarding the placement of the combined resistance Rcr, particularly in relation to R5 and R3. Clarifications indicate that Rcr can be positioned at various points in the circuit without affecting its parallel relationship with R5. The conversation emphasizes understanding the arrangement of resistances to accurately compute the total resistance.
spoonthrower
Messages
36
Reaction score
0
AxxxxxxR1xxxxxxxxxxxxxR2xxxxxxxxR3xxxxxxx
ooooooooooooooXoooooooooXooooooooooox
ooooooooooooooXoooooooooXooooooooooox
ooooooooooooooR4ooooooooR5oooooooooR6
ooooooooooooooXoooooooooXooooooooooox
BxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxR7xxxxxx

For the circuit above the x's represent the wire and R represents the resistance. o's just fill in open space - ignore them. Find the equivalent resistance between points A and B.
R1=5 ohms
R2=4 ohms
R3=6 ohms
R4=10 ohms
R5=8 ohms
R6=7 ohms
R7=3 ohms

So i need to find the total resistance in the circuit. I know that R4,5, and 6 are in parallel so their resistance should add up like 1/(1/10+1/8+1/7) to give 2.72 ohms. so the rest of the resistances are in series so you just add them all up right? when i add them all together with 2.72 ohms, I am not getting the right answer. What am i doing wrong? please help. thanks.
 
Last edited:
Physics news on Phys.org
R3, 6 and 7 are in series. Start by trying to calculate their combined resistance first.
 
Adding to what andrevdh suggests, a good place to start is the at the end of the circuit. After you've found a combined resistance Rcr for those andrevdh described, you can then work your way towards the right. Notice R5 is now in parallel with your new combined resistance Rcr; you can combine these two to get a new combined resistance. Can you see the process?
 
Last edited:
i do not see how Rcr is in parallel with R5 at all. If it is in parallel with R5 that means that when u add Rcr up, it would have to be at the R6 location. Why can't u place the new Rcr at the R3 location. i am really confused.
 
spoonthrower said:
i do not see how Rcr is in parallel with R5 at all. If it is in parallel with R5 that means that when u add Rcr up, it would have to be at the R6 location. Why can't u place the new Rcr at the R3 location. i am really confused.

You could put the combined resistance Rcr (combined from R3, R6 and R7) anywhere in the branch at the extreme right. You could place it wheer R3 is or wher R6 is or where R7 is. In *all* three cases, Rcr is still in parallel with R5
 
got it. thank you so much.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top