CBSE Class 12-science Answered

This circuit  diagram is as shown in figure. Let I₁ and I₂ be current given out by each cell.
Applying Kirchoff's second law to closed mesh PQRS we have,
2I₁ + 2I₁+ (I₁ + I₂)8 + 3I₁ = 1.5
or 15 I₁ + 8 I₂ = 1.5 ........(1)
Applying kirchhoff's law to closed mesh RSNMR, we have
8(I₁ + I₂) + 3I₂ + 1I₂ + 2I₂ = 2
or 8I₁ + 14I₂ = 2
or 4I₁ + 7I₂ = 1.................(2)
Multiplying (1)  by 7 and (2) by 8, we have
105I₁ + 56I₂ = 10.5 ...........(3)
32I₁ + 56I₂ = 8 ...........(4)
Subtracting, (4) from (3) we get, 73I₁ = 2.5
So, I₁  = 2.5/73 = 5/146 amp.
Putting this value of I₁ in (2),
4 x 5/146 + 7I₂ = 1
or 7I₂ = 1 - 20/146 = 126/146
I₂ = (126/146)/7 = 18/146 amp.
Thus, current through 8 ohms resistance
= I₁ + I₂ = 5/146 + 18/146
= 23/146 amp.
which is required value of current.
Also P.D across 8 ohms wire
= current x resistance
= (23/146) x 8 = 184/146 volts
= 1.26 volts