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The circuit in Fig. shows two cells connected in opposition toeach other. Cell epsi_1, is of emf 6 V and internal resistance 2Omega and the cell epsi_2is of emf 4 V and internal resistance 8 Omega. Find the potential difference between the points A and B. |
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Answer» Solution :Since two cells are connected in opposition, hence net emf `EPSI = epsi_1 - epsi_2 = 6-4 = 2 V `and total resistance `r = r_1 + r_2 = 2 + 8 = 10 Omega` ` therefore` Circuit current`I = epsi/r = 2/10 = 0.2 A` ` therefore ` Potential difference between the POINTS A and `B = |V_A-V_B | = epsi_2 + r_2I = 4 + 8 xx 0.2 = 5.6 V` Of course point B is at higher potential than the point A. |
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