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NCERT Exemplar · Q22

Q.Two cells are connected in parallel in opposition: one cell has emf 10 V10\ \text{V} and internal resistance 10 Ω10\ \Omega, the other has emf 2 V2\ \text{V} and internal resistance 5 Ω5\ \Omega, and they are joined so that the positive terminal of the 10 V10\ \text{V} cell is connected to the negative terminal of the 2 V2\ \text{V} cell (their other two terminals also being joined). Find the effective (equivalent) emf and the effective internal resistance of this parallel combination.

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Two cells in parallel behave like one equivalent cell. Its emf is the average of the individual emfs weighted by their conductances 1/r1/r, and its internal resistance is the parallel combination of the two internal resistances. Because the 2 V2\ \text{V} cell is connected in opposition, its emf enters with a minus sign, giving Eeff=2 VE_{eff} = 2\ \text{V} and reff=10/3 Ωr_{eff} = 10/3\ \Omega.

Concept

For two cells in parallel between the same pair of terminals, the equivalent-source (Millman) results are

Eeff=E1r1+E2r21r1+1r2,reff=r1r2r1+r2,E_{eff} = \frac{\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}}{\dfrac{1}{r_1} + \dfrac{1}{r_2}}, \qquad r_{eff} = \frac{r_1 r_2}{r_1 + r_2},

where each emf carries a sign according to its polarity in the circuit.

Assign values with signs

Take the 10 V10\ \text{V} cell as positive: E1=+10 VE_1 = +10\ \text{V}, r1=10 Ωr_1 = 10\ \Omega. The 2 V2\ \text{V} cell opposes it (positive of 10 V10\ \text{V} joined to negative of 2 V2\ \text{V}), so E2=−2 VE_2 = -2\ \text{V}, r2=5 Ωr_2 = 5\ \Omega.

Effective emf …

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