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## Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12

Question

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 2.00 A. When the resistors are connected in parallel to the battery, the total current from the battery is 10.2 A. Determine the two resistances.

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Physics
2 months
2021-08-28T08:03:48+00:00
2021-08-28T08:03:48+00:00 1 Answers
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## Answers ( )

Answer:

1.61ohms and 4.39ohms

Explanation:

According to ohm’s law which States that the current (I) passing through a metallic conductor at constant temperature is directly proportional to the potential difference (V) across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the current

Rt is the effective resistance

Let the resistances be R and r

When the resistors are connected in series to a 12.0-V battery and the current from the battery is 2.00 A, the equation becomes;

12 = 2(R+r)

Rt = R+r (connection in series)

6 = R+r …(1)

If the resistors are connected in parallel to the battery and the total current from the battery is 10.2 A, the equation will become;

12 = 10.2(1/R+1/r)

Since 1/Rt = 1/R+1/r (parallel connection)

Rt = R×r/R+r

12 = 10.2(Rr/R+r)

12(R+r) = 10.2Rr … (2)

Solving equation 1 and 2 simultaneously to get the resistances. From (1), R = 6-r…(3)

Substituting equation 3 into 2 we have;

12{(6-r)+r} = 10.2(6-r)r

12(6-r+r) = 10.2(6r-r²)

72 = 10.2(6r-r²)

36 = 5.1(6r-r²)

36 = 30.6r-5.1r²

5.1r²-30.6r +36 =

r = 30.6±√30.6²-4(5.1)(36)/2(5.1)

r = 30.6±√936.36-734.4/10.2

r = 30.6±√201.96/10.2

r = 30.6±14.2/10.2

r = 44.8/10.2 and r = 16.4/10.2

r = 4.39 and 1.61ohms

Since R+r = 6

R+1.61 = 6

R = 6-1.61

R = 4.39ohms

Therefore the resistances are 1.61ohms and 4.39ohms