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NCERT Exemplar · Q18

Q.Two cells are connected in opposition to each other, forming a single closed loop with no external resistor. Cell E1E_1 has emf 6 V6\ \text{V} and internal resistance 2 Ω2\ \Omega; cell E2E_2 has emf 4 V4\ \text{V} and internal resistance 8 Ω8\ \Omega. Their emfs act against each other around the loop, and AA and BB are the two junction points at which the two cells meet. Find the potential difference between the points AA and BB.

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Because the two cells are connected in opposition, only their difference in emf drives current. The net emf 6−4=2 V6-4 = 2\ \text{V} pushes a current of 0.2 A0.2\ \text{A} through the total internal resistance 2+8=10 Ω2+8 = 10\ \Omega. Evaluating the terminal voltage of either cell gives the potential difference across AA and BB as 5.6 V5.6\ \text{V}.

Concept

With no external resistor, the two cells form one loop. Being in opposition, the effective emf is E1−E2E_1 - E_2 and the total resistance is the sum of the internal resistances.

Step 1 — loop current

I=E1−E2r1+r2=6−42+8=210=0.2 A.I = \frac{E_1 - E_2}{r_1 + r_2} = \frac{6 - 4}{2 + 8} = \frac{2}{10} = 0.2\ \text{A}.

The current is driven by the stronger cell E1E_1; the weaker cell E2E_2 is being charged (current is forced through it against its emf).

Step 2 — potential difference between A and B …

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