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www.GOALias.blogspot.comPhysicsIf we wish to replace the combination by a single cell between A andC of emf ε eqand internal resistance r eq, we would haveV AC= ε eq– I r eq(3.63)Comparing the last two equations, we getε eq= ε 1+ ε 2(3.64)and r eq= r 1+ r 2(3.65)In Fig.3.20, we had connected the negative electrode of the first to thepositive electrode of the second. If instead we connect the two negatives,Eq. (3.61) would change to V BC= –ε 2–Ir 2and we will getε eq= ε 1– ε 2(ε 1> ε 2) (3.66)The rule for series combination clearly can be extended to any numberof cells:(i)FIGURE 3.21 Two cells inparallel. For connectionsacross A and C, thecombination can bereplaced by one cell of emfε eqand internal resistancesr eqwhose values are given inEqs. (3.64) and (3.65).The equivalent emf of a series combination of n cells is just the sum oftheir individual emf’s, and(ii) The equivalent internal resistance of a series combination of n cells isjust the sum of their internal resistances.This is so, when the current leaves each cell from the positiveelectrode. If in the combination, the current leaves any cell fromthe negative electrode, the emf of the cell enters the expressionfor ε eqwith a negative sign, as in Eq. (3.66).Next, consider a parallel combination of the cells (Fig. 3.21).I 1and I 2are the currents leaving the positive electrodes of thecells. At the point B 1, I 1and I 2flow in whereas the current I flowsout. Since as much charge flows in as out, we haveI = I 1+ I 2(3.67)Let V (B 1) and V (B 2) be the potentials at B 1and B 2, respectively.Then, considering the first cell, the potential difference across itsterminals is V (B 1) – V (B 2). Hence, from Eq. (3.57)( ) ( )V ≡ V B1 – V B2 = ε1 – I1r1(3.68)Points B 1and B 2are connected exactly similarly to the secondcell. Hence considering the second cell, we also have( ) ( )V ≡ V B1 – V B2 = ε2 – I2r2(3.69)Combining the last three equationsI = I1 + I2ε – V ε – V ⎛ε ε ⎞ ⎛1 1⎞– Vr r ⎜⎝r r ⎟⎠⎜⎝r r ⎟⎠1 2 1 2= + = + +1 2 1 2 1 2Hence, V is given by,(3.70)114ε r 1 2+ε r 2 1 1 2V = – Irrr + r r + r1 2 1 2(3.71)If we want to replace the combination by a single cell, between B 1andB 2, of emf ε eqand internal resistance r eq, we would haveV = ε eq– I r eq(3.72)