Question

when two resistors are connected in series with a cell of emf 2v and negligible internal resistance,a current of 2/5A flows in the circuit. when the resistors are connected in parallel the main current is 5/3A . calculate the resistances.

Answer 1

In series
V=IR
I=V/R
I=2/0.4
I=5A
In parallel
I=2/1.666
I=1.199A

Answer 2

Answer:  R1 = 2 ohms and R2 = 3 ohms

Explanation:

V = 2 V

I1 = 2/5 A

R series = R1 + R2

            = V / I1

            = 5 ohms               ..............(1)

I2 = 5/3 A

1/ R parallel  = 1/ R1 + 1/ R2

   R parallel  =   R1 x R2 / R1 + R2

      V / I2 = R1 x R2 / 5

      R1 x R2 / 5  =  2 / 5/3

      R1 x R2  = 6               ..............(2)

Comparing equations (1) and (2)

we get R1 = 2 ohms and R2 = 3 ohms