Question

Two resistors are connected in series with 5v battery of negligible internal resistance.A current of 2A flows through each resistor. If they are connected in parallel with the same battery a current of 25/3A flows through combination calculate the value of each resistance



i want with proof​

Answer 1

Explanation:

let,

v=ir

first in series Req=r1+r2

so 5=2(r1+r2)

then r1+r2=5/2

then in parallel

Req=r1r2/r1+r2

so v=ir

5=25/3*(r1r2/r1+r2)

15(r1+r2)=25r1r2

15*5/2=25r1r2

75/50=r1r2

3/2=r1r2

let form a quadratic equation

x^2-x(r1+r2)+r1r2

x^2-x(5/2)+3/2=0

multiply by 2

2x^2-5x+3=0

so 2x^2-3x-2x +3

so r1=3/2 and r2=1