Two cells with EMFs E₁ = 2 V and E₂ = 4 V , and internal resistances r₁ = 1 and r₂ = 2 respectively, are connected in parallel across an external resistor R . The value of R is chosen such that the power dissipated in it is maximum. The maximum power dissipated in R is:

Step-by-step solution

First, we find the equivalent EMF ( E_ eq ) and equivalent internal resistance ( r_ eq ) of the parallel combination of the two cells. The equivalent internal resistance is: r_ eq = r₁ r₂ r₁ + r₂ = 1 2 1 + 2 = 2 3 The equivalent EMF is given by: E_ eq = E₁ r₁ + E₂ r₂ 1 r₁ + 1 r₂ = 2 1 + 4 2 1 1 + 1 2 = 2 + 2 3 2 = 4 3 2 = 8 3 V According to the Maximum Power Transfer Theorem, the power dissipated in the external resistor R is maximum when R = r_ eq = 2 3 . The maximum power P_ max is given by: P_ max = E_ eq ^2 4R