Two resistors of resistances R_1 and R_2, with R_2 > R_1, are connected to a voltage source with voltage V_0. When the resistors are connected in series, the current is I_s. When the resistors are connected in parallel, the current I_p from the source is equal to 10I_s.
Let r be the ratio R_1/R_2. Find r.
Round your answer to the nearest thousandth.
Any help is appreciated, Thanks!
2 Answers
-
Series: I = Is = V/(R1+R2)
parallel, R = R1*R2/(R1+R2)
Ip = 10*Is = V/R = V(R1+R2)/R1*R2
10V/(R1+R2) = V(R1+R2)/R1*R2
10/(R1+R2) = (R1+R2)/R1*R2
10*R1*R2 = (R1+R2)(R1+R2)
10*R1*R2 = 2R1R2 + R1² + R2²
8R1R2 = R1² + R2²
r = R1/R2
R1 = rR2
8R1R2 = R1² + R2²
8rR2R2 = (rR2)² + R2²
8rR2² = (rR2)² + R2²
R2² (8r – r² – 1) = 0
assuming R2 is not zero
r² – 8r + 1 = 0
quadratic equation:
to solve ax² + by + c = 0
x = [-b ±√(b²-4ac)] / 2a
r = [8 ±√(64-4)] / 2
r = [8 ± 7.7460] / 2
r = 0.127
needs checking
.
-
one of the resistors is 6 ohms the other is 4 ohms
