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
