Question

When a 36.0-V emf device is placed across two resistors in series, a current of 10.0 A is flowing in each of the resistors. When the same emf device is placed across the same two resistors in parallel, the current through the emf device is 55.0 A. What is the magnitude of the larger of the two resistances?

Answer 1

Let the resistors be R1 and R2

In series,

I = V / (R1+R2)

10 = 36 / (R1+R2)

R1+R2 = 3.6

R1 = 3.6 - R2

In parallel,

net Resistance = R1*R2/R1+R2

= (3.6-R2)*R2/3.6

Now use:

I = V/net R

55 = 36 / (3.6-R2)*R2/3.6

55 = 129.6 / (3.6-R2)*R2

(3.6-R2)*R2 = 2.356

3.6*R2 - R2^2 = 2.356

R2^2 - 3.6*R2 + 2.356 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = -3.6

c = 2.356

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.536

roots are :

R2 = 2.74 and R2 = 0.8598

These are two resistors value

Answer: 2.74 ohm