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Two cells A and B of e.m.f.2 V and 1.5 V respectively, are connected as shown in figure through an external resistance 10 Omega The internal resistance of each cell is 5 Omega. The potential difference E_A and E_B across the terminals of the cells A and B respectively are |
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Answer» `E_A = 2.0v,E_B = 1.5 V` `2-1.5 -5I-10 I-5I =0` `implies0.5= 20I impliesI=(0.5 )/(20)=(1)/(40 )A ` terminalpotentialdiffernceacrossthe cell A ` E_A= 2-Ir=2 -(1)/(40 ) xx 5 =1.875V.` terminalpotentialdifference acrossthe cell B ` E_B= 1.5+Ir= 1.5+(1)/(40)xx 5 = 1.875V.` terminalpotentialdiffernceacrossthe cellB. ` E_B= 1.5 +Ir = 1.5+(1)/(40) xx5 =1.625 V.` 
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