when a current of 0.5 ampere is passed through two resistances in series the potential difference between the two ends of the series arrangement is 12.5v on connecting them in parallel and passing the current of 1.5 ampere the potential difference between their ends is 6 volt calculate the two resistances
Answers
Answered by
18
in the first case Rs= V/I
Rs= 12.5/.5= 25 ohm i.e. R1+R2=25
in the second case Rp=V/I
Rp= 6/1.5=4 ohm i.e. 1/R1+1/R2=1/4 ; R1*R2/(R1+R2)=4
Substituting the value of R1+R2=25 from above we get R1*R2=4*25=100
So R1+R2=25
and R1*R2=100 So the probable resistances are 20 and 5 ohms
Rs= 12.5/.5= 25 ohm i.e. R1+R2=25
in the second case Rp=V/I
Rp= 6/1.5=4 ohm i.e. 1/R1+1/R2=1/4 ; R1*R2/(R1+R2)=4
Substituting the value of R1+R2=25 from above we get R1*R2=4*25=100
So R1+R2=25
and R1*R2=100 So the probable resistances are 20 and 5 ohms
Similar questions