Two cells of voltage 10 V and 2 V and internal resistances 10 `Omega and 5 Omega` respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2 V battery (figure). Find the effective voltage and effective resistance of the combination.

Explanation

By using Kirchhoff.s first law at point A,
`I_(1) = I + I_(2 ) " "` ... (1)
By using Kirchhoff.s second law for loop EDCFE,
`-IR - 10I_(1) + 10 = 0`
`therefore IR + 10I_(1) = 10 " " ` ... (2)
For ADCBA loop,
`- IR + 5 I_(2 ) - 2 = 0 `
` - IR + 5 (I_(1) - I) = 2 `
` therefore -IR + 5I_(1) -5 I = 2 `
`therefore - 2IR + 10 I_(1) - 10 I = 4 ` ... (3)
Subtracting equation (3) from equation (2),
`therefore 3 IR + 10 I = 6 `
`therefore IR + (10 xx 1)/(3 ) = 2 `
`therefore I (R + (10)/(3)) = 2 " " ` .... (4)
From Ohm.s law,
`I (R + R_(eq)) = V_(eq) " " `... (5) `[ because IR = V] `
Comparing equation (4) and (5) ,
`R_(eq) = (10)/(3) Omega and V_(eq) = 2 V `
Here , `R_(eq)` is equivalent resistance of 10 `Omega and 5 Omega`. Hence, equivalent circuit will be as shonw below.