Resistors

Two parallel resistors connected in series give the resulting resistance $R1Ω and $R2Ω. Determine the resistance of these resistors.

Correct answer:

R1 =  168 Ω
R2 =  56 Ω

Step-by-step explanation:

S=224 Ω P=42 Ω  R1+R2 = 224 Ω R11+R21 =   42  Ω1 R2 = 224  R1  R11+ 224  R11 =   42  Ω1  R2224 R+224 42=0  R2224 R+224 42=0 R2224R+9408=0  a=1;b=224;c=9408 D=b24ac=2242419408=12544 D>0  R1,2=2ab±D=2224±12544 R1,2=2224±112 R1,2=112±56 R1=168=168 Ω R2=56   Verifying Solution:   S2=R1+R2=168+56=224 Ω P2=R1+R2R1 R2=168+56168 56=42 Ω

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