Question

A circuit consisting of three resistances 12, 18 and 36 respectively joined in parallel is connected in series with a fourth resistance. The whole circuit is supplied at 60 V and it is found that power dissipated in 12 ohm resistance is 36W. Determine the value of fourth resistance and the total power dissipated in the group.

Answer 1

Answer:

Explanation:

Given A circuit consisting of three resistances 12, 18 and 36 respectively joined in parallel is connected in series with a fourth resistance. The whole circuit is supplied at 60 V and it is found that power dissipated in 12 ohm resistance is 36 W. Determine the value of fourth resistance and the total power dissipated in the group.

We know that power = v^2 / R

So p = v^2 / R

  36 = v^2 / 12

V^2 = 12 x 36

V^2 = 432

So v = 20.78 v

This will be voltage across R1, R2 and R3

Now I 1 = V1 / R1 = 20.78 / 12 = 1.73

    I 2 = 20.78 / 18 = 1.15

     I 3 = 20.78 / 36 = 0.577

So I 4 = I 1 + I 2 + I 3 = 1.73 + 1.15 + 0.577 = 3.457 A

Now R4 = V4 / I 4

               = 60 – 20.78 / 3.457

             = 39.22 / 3.457

          = 11.34 ohms

Now power  p = E x I4

                     = 60 x 3.457

                  = 207.42 watts