Resistors

Two resistors connected in series give the resulting resistance 216Ω and 48Ω in parallel. Determine the resistance of these resistors.

Result

R1 =  72 Ω
R2 =  144 Ω

Solution:

 R1+R2=216 1R1+1R2=148 R2=216−R1  1R1+1216−R1=148  R12−216R1+216⋅48=0 R12−216R1+10368=0 (R1−72)⋅(R1−144)=0  R1=72Ω R2=144Ω  \ \\ R_1+R_2 = 216 \ \\ \dfrac{1}{R_1}+\dfrac{1}{R_2} = \dfrac{1}{ 48} \ \\ R_2 = 216 - R_1 \ \\ \ \\ \dfrac{1}{R_1}+\dfrac{1}{ 216 - R_1} = \dfrac{1}{ 48} \ \\ \ \\ R_1^2-216 R_1 + 216\cdot 48 = 0 \ \\ R_1^2-216 R_1 + 10368 = 0 \ \\ (R_1-72)\cdot (R_1-144) =0 \ \\ \ \\ R_1 = 72 \Omega \ \\ R_2 = 144 \Omega \ \\

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