Answer:- R1=R, i1=5A
R2=10+R10R, i2=6A
Since probable constant,
i1R1=i2R2
⇒5×R=10+R6×10R
⇒(10+R)5=60
⇒5R=10⇒R=2Ω
An ideal battery sends a current of 5A in a resistor. When another resistor of value 10Ω is connected in parallel, the current through the battery is increased to 6A. find the resistance of the first resistor.
Posted: 19-08-2022