Question

Two cells of emf \(E_1\) and \(E_2\) with internal resistances \(r_1\) and \(r_2\) respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.

Hint:

For two cells connected in parallel, \[ E_{\text{eq}} = \frac{E_1r_2+E_2r_1} {r_1+r_2} \] and \[ r_{\text{eq}} = \frac{r_1r_2}{r_1+r_2}. \] If both cells have equal internal resistances, the equivalent emf becomes the arithmetic mean of the two emfs.

Answer 1

Concept: When two cells are connected in parallel, the combination can be replaced by a single equivalent cell having an equivalent emf \(E_{\text{eq}}\) and an equivalent internal resistance \(r_{\text{eq}}\). The equivalent cell should supply the same current to any external circuit as the original combination. To determine the equivalent emf and internal resistance, we make use of Kirchhoff's laws and Thevenin's equivalent concept.

Step 1: Consider the parallel combination of the two cells. Let the positive terminals of the cells be connected together and the negative terminals also be connected together. Suppose the open-circuit terminal voltage of the combination is \(V\). Since no external current is drawn under open-circuit condition, the algebraic sum of currents in the two branches must be zero. \[ I_1+I_2=0 \] or \[ I_1=-I_2. \]

Step 2: Write the terminal voltage across each cell. For the first cell, \[ V=E_1-I_1r_1 \] and for the second cell, \[ V=E_2-I_2r_2. \] Since \[ I_2=-I_1, \] we obtain \[ E_1-I_1r_1 = E_2+I_1r_2. \] Therefore, \[ I_1(r_1+r_2) = E_1-E_2. \] Hence, \[ I_1 = \frac{E_1-E_2}{r_1+r_2}. \]

Step 3: Determine the equivalent emf. Substituting \(I_1\) into \[ V=E_1-I_1r_1, \] we get \[ V = E_1-\frac{r_1(E_1-E_2)}{r_1+r_2}. \] Taking the LCM, \[ V = \frac{E_1(r_1+r_2)-r_1(E_1-E_2)} {r_1+r_2}. \] Simplifying, \[ V = \frac{E_1r_2+E_2r_1} {r_1+r_2}. \] This open-circuit voltage is the equivalent emf. Therefore, \[ \boxed{ E_{\text{eq}} = \frac{E_1r_2+E_2r_1} {r_1+r_2} } \]

Step 4: Determine the equivalent internal resistance. To find internal resistance, replace each ideal emf source by a short circuit. The internal resistances \(r_1\) and \(r_2\) then appear in parallel. Hence, \[ r_{\text{eq}} = \frac{r_1r_2}{r_1+r_2}. \] Therefore, \[ \boxed{ r_{\text{eq}} = \frac{r_1r_2}{r_1+r_2} } \] Final Result: Equivalent emf: \[ \boxed{ E_{\text{eq}} = \frac{E_1r_2+E_2r_1} {r_1+r_2} } \] Equivalent internal resistance: \[ \boxed{ r_{\text{eq}} = \frac{r_1r_2}{r_1+r_2} } \]