Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.78 A. When the resistors are connected in parallel to the battery, the total current from the battery is 11.3 A. Determine the two resistances.

R(smaller) = 1.3 Ω  and  R(larger) = 5.4 Ω

Explanation:

Ohm's Law states that:

V = IR

R = V/I

where,

R = Resistance

V = Potential Difference

I = Current

Therefore, for series connection:

Rs = Vs/Is

where,

Rs = Resistance when connected in series = R(smaller) + R(larger)

Vs = Potential Difference when connected in series = 12 V

Is = Current when connected in series = 1.78 A

Therefore,

R(smaller) + R(larger) = 12 V/1.78 A

R(smaller) + R(larger) = 6.74 Ω   equation 1

R(smaller) = 6.74 Ω - R(larger)     equation 2

Therefore, for series connection:

Rp = Vp/Ip

where,

Rp = Resistance when connected in parallel = [1/R(smaller) + 1/R(larger)]⁻¹

Rp = [{R(smaller) + R(larger)}/{R(smaller).R(larger)]⁻¹

Rp = R(smaller).R(larger)/[R(smaller) + R(larger)]

Vp = Potential Difference when connected in parallel = 12 V

Ip = Current when connected in parallel = 11.3 A

Therefore,

R(smaller).R(larger)/[R(smaller) + R(larger)] = 12 V/11.3 A

using equation 1 and equation 2, we get:

[6.74 Ω - R(larger)].R(larger)/6.74 Ω = 1.06 Ω

6.74 R(larger) - R(larger)² = (6.74)(1.06)

R(larger)² - 6.74 R(larger) + 7.16 = 0

solving this quadratic equation we get:

R(larger) = 5.4 Ω (OR) R(larger) = 1.3 Ω

using these values in equation 2, we get:

R(smaller) = 1.3 Ω (OR) R(smaller) = 5.4 Ω

Since, it is given in the question that R(smaller)<R(larger).

Therefore, the correct answers will be:

R(smaller) = 1.3 Ω  and  R(larger) = 5.4 Ω

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explanation:

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