# Resistance problem- combination circuit

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.

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Phlogistonian
R2 and R5 are in series with R3 and R4. They should be included in the calculation of RS.

drdizzard
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 consistant. Anywho, thanks!

drdizzard
That's correct the total resistance of the circuit is 57 ohms. I am glad I was able to help.