# 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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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 consistant. Anywho, thanks!

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