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 1.99 A. When the resistors are connected in parallel to the battery, the total current from the battery is 10.7 A. Determine the two resistances.

Answer 1

R(smaller) =r
R(larger) = R
in series:
1.99(R+r)=12 . ... ..(1)
in parallel:
1/R +1/r=1/R₀
R₀=rR/(R+r)
10.7 R₀=12
10.7•rR/(R+r)=12 …..(2)
Multiply (1) by (2)
1.99• (R+r)•10.7•rR/(R=r) =144
r=6.76/R …… (3)
From (1)
R+r=6.03 ……(4)
Substitute (30 in (4)
6.76/R + R =6.03
R²-6.03R+6.76 =0
R = 3.015±sqrt(9.09-6.76) =
=3.015±1.53
Two roots:
R₁=4.545 => r₁=6.76/R₁=1.49
R₂=2.485 => r₂=6.76/R₂=2.72
We take the first pair as R>r
Ans. : R₁=4.545 Ω, r₁=1.49 Ω