Perform a series-parallel reduction on a network: 6 Ω in series with parallel of 12 Ω and 3 Ω; then in series with a 9 Ω resistor and an 18 V source. Find the equivalent resistance of the parallel pair (12 Ω and 3 Ω).

Perform a series-parallel reduction on a network: 6 Ω in series with parallel of 12 Ω and 3 Ω; then in series with a 9 Ω resistor and an 18 V source. Find the equivalent resistance of the parallel pair (12 Ω and 3 Ω).

• A. 2.4 Ω
• B. 1.8 Ω
• C. 4.0 Ω
• D. 3.2 Ω

Answer: (A)

When resistors are in parallel, you add their conductances, not their resistances. The conductance is the reciprocal of resistance, so 1/R = 1/R1 + 1/R2. For 12 Ω and 3 Ω: 1/R = 1/12 + 1/3 = 1/12 + 4/12 = 5/12 R = 1 / (5/12) = 12/5 = 2.4 Ω So the equivalent resistance of the parallel pair is 2.4 Ω. (If you continued, this would then be in series with 6 Ω and 9 Ω, giving a total of 17.4 Ω.)

When resistors are in parallel, you add their conductances, not their resistances. The conductance is the reciprocal of resistance, so 1/R = 1/R1 + 1/R2.

For 12 Ω and 3 Ω:

1/R = 1/12 + 1/3 = 1/12 + 4/12 = 5/12

R = 1 / (5/12) = 12/5 = 2.4 Ω

So the equivalent resistance of the parallel pair is 2.4 Ω. (If you continued, this would then be in series with 6 Ω and 9 Ω, giving a total of 17.4 Ω.)