Question

Two resistors connected in series have an equivalent resistance of 699.9 Ω. When they are connected in parallel, their equivalent resistance is 124.3 Ω. Find the resistance of each resistor.

Ω (small resistance)

Ω (large resistance)

Answer 1

Answer 2

SOLUTION :

2 resistors, R1 and R2 connected in series = 699.9 ohms

=> R1 + R2 = 699.9 ohms

=> R2 = 699.9 - R1

When they are connected in parallel, the equivalent is = 124.3 ohms 

=> 1/ [1/R1 + 1/R2] = 124.3 ohms

=> R1 R2 / [R1 + R2] = 124. 3 ohms

Substituting value of R2 = 699.9 - R1 and R1 + R2 = 699.9 :

=> R1(699.9 - R1) / 699.9 = 124.3 

=> 699.9 R1 - R1^2 = 124.3 * 699.9

=> R1^2 - 699.9 R1 + 86997.57 = 0 

=> R1 = 699.9/2 +/- sqrt(699.9^2 - 4(1)(86997.57)) / 2

=> R1 = 349.95 +/- 188.33

=> R1 = 161.62, 538 28 

So, if R1 = 161.62, R2 = 699.9 - 161.62 = 538.28 and vice-versa.

Hence, 

R1 = 161.62 ohms and R2 = 538.28 ohms 

OR

R1 = 538.28 ohms and R2 = 161.62 ohms.

It means one restore is 161.62 ohms and the other resistor is 538.28 ohms 

(ANSWER)