Resistance problem- combination circuit

Click For Summary

Homework Help Overview

The discussion revolves around calculating the total resistance in a combination circuit involving multiple resistors connected in series and parallel. The original poster presents their approach using Ohm's Law and formulas for series and parallel resistances.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply Ohm's Law and the formulas for series and parallel resistors to find the total resistance. Some participants question the inclusion of certain resistors in the calculations, suggesting that R2 and R5 should be considered in the series calculation with R3 and R4. Others propose alternative configurations and calculations for the resistors.

Discussion Status

There is an ongoing exploration of different interpretations of the circuit configuration. Some participants provide corrections and alternative calculations, while others express uncertainty about the original approach. Guidance has been offered regarding the arrangement of the resistors, but no consensus has been reached on the final total resistance.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. The discussion includes corrections and clarifications regarding the arrangement of resistors in the circuit.

adv
Messages
6
Reaction score
0
Question 3. Find the total resistance of the following circuit.

Formulas:

R= V/I

If the # of resistors connected in series is n, the equiv. resistor is given by
Rs = R1 + R2 +... + Rn

likewise with resistors connected in parallel,
1/Rp = 1/R1 + 1/R2 +... + 1/Rn

Given:
V0 = 120 V
R1 = 20 Ω
R2 = 22 Ω
R3 = 30 Ω
R4 = 27 Ω
R5 = 40 Ω
R6 = 20 Ω
R7 = 20 Ω

My attempt:

Applying Ohm’s Law to the entire circuit:

I0 = V0 / RT , where RT is the total resistance in the circuit

But since R3 and R4 are connected in series,

RS = R3 + R4
= 30 Ω + 27 Ω
= 57 Ω

Then RT1 and R7 are connected in parallel:

1/RP = 1/RS + 1/R7
= 1/57 + 1/20
= 20/1140 + 57/1140
= 77/1140 Ω
RP = 14.8 Ω
= 15 Ω

Solve five resistors R1, R2, RP, R5 and R6. These resistors are connected in series

RT = R1 + R2 + R3 + R4 + R5 + R6
= 20 Ω + 22 Ω + 15 Ω + 40 Ω + 20 Ω
= 117 Ω

Therefore, the total resistance of the circuit is 117 Ω.

I'm not sure I took the right approach. Any help would be appreciated. Thanks.
 

Attachments

Physics news on Phys.org
R2 and R5 are in series with R3 and R4. They should be included in the calculation of RS.
 
R2, R3, R4, and R5 are all in series

R8=R2+R3+R4+R5

R8 is parrallel to R7

1/R9=1/R8 + 1/R7

R9 is in series with R1 and R6

RTotal=R1+R9+R6
 
Here is the correction:

Since R2, R3, R4 and R5 are connected in series,

R8 = R2 + R3 + R4 + R5
= 22 Ω + 30 Ω + 27 Ω + 40 Ω
= 119 Ω

Then R7 and R8 are connected in parallel:

1/R9 = 1/R7 + 1/R8
= 1/20 + 1/119
= 119/2380 + 20/2380
= 139/2380 Ω
R9 = 17.1 Ω
= 17 Ω

Solve three resistors R1, R9 and R6. These resistors are connected in series

RT = R1 + R9 + R6
= 20 Ω + 17 Ω + 20 Ω
= 57 Ω

Therefore the total resistance of the circuit is 57 Ω.

I think this is correct now. Thank you Phlogistonian and Drdizzard. BTW, the relabeling of R8 and R9 was less confusing than Rs and Rp. I guess it was more consistent. Anywho, thanks!
 
That's correct the total resistance of the circuit is 57 ohms. I am glad I was able to help.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K