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Two cells of same emf epsibut internal resistances r_1 and r_2are connected in series to an externalresistor R. What should be the value of R so that the potential difference across the terminals of the first cell becomes zero. |
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Answer» Solution : Here net emf of combination = `2epsi`and net resistance of the circuit = `r_1 + r_2 + R` CURRENT flowing in the circuit `I = (2 EPSI)/(r_1 + r_2 + R)` As potential difference across FIRST CELL `V_1 = epsi - Ir_1= 0, ` hence `epsi - (2epsi)/((r_1 + r_2 + R)) r_1 = 0 ` or`epsi[r_1 + r_2 +R - 2r_1]= 0` ` rArr R = r_1 - R_2 ` |
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