Question

An ideal battery sends a current of 5A in a resistor.When another resistor of value 10 ohm is connected in parallel,the current through the battery is increased to 6A.Find the resistance of the first resistor. pls answer

Answer 1

here
I=6A=. , R1= ₹
R2=10ohm
the resistor connected in parallel (it is given)
now,
I=1/R1+1/R2
6 =1/R1+1/10
6=10+R1/10×R1. (by crosss multiple)
6×10R1=10+R1
60R1=10+R1
60R1-R1=10
59R1=10
R1=10/59
R1=5.9 ohm of the first resistor

Answer 2

r=? R1 =10

R(eq)=R1r/R1+r

=10r/10+r

Since potential is same,becoz battery gone a be same

5*r=6*10r/10+r

r=2 ohm