Two cells A and B of emf 2 V and 1.5 V respectively, are connected as shown in figure through an external resistance 10Ω The internal resistance of each cell is 5Ω . The potential difference E A and

Question

Two cells A and B of emf 2 V and 1.5 V respectively, are connected as shown in figure through an external resistance 10Ω The internal resistance of each cell is 5Ω . The potential difference E A ​ and E B ​ across the terminals of the cells A and B respectively are

TestHub · Accepted Answer

The figure can be redrawn as, The current through the circuit i ​ = effective resistance net emf ​ = 5 + 5 + 10 2 − 1.5 ​ = 20 0.5 ​ = 40 1 ​ = 0.025 A ​ The terminal potential difference of the batteries and V A ​ V B ​ ​ = ε A ​ − i I A ​ = 2 − 0.025 × 5 = 2 − 0.0125 = 1.875 V = ε B ​ + i I B ​ ​ = 1.5 + 0.025 × 5 = 1 , 5 + 0.0125 = 1.625 V

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