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When a current drawn from a battery is 0.5 A, its terminal potential difference is 20 V. And when current drawn from it is 2.0 A , the terminal voltage reduces to 16 V . Find out e.m.f and internal resistance of the battery. |
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Answer» Solution :Let E and R be the e.m.f and internal resitance of battery. When a current I ampere is drawn from it, then POTENTIAL drop across internal resitance or inside the CELL is = Ir. Then `V= E- Ir, I= 0.5 A , V= 20 ` VOLT, we have `20 = E- 0.5 r "".....(i)` `I = 2 A , V= 16 ` Volt, we have `16- E- 2r "".....(ii)` From eqs (i) and (ii) `2E - r = 40 and E= 2r = 16` So solving we get `E= 21.3 V, r = 2.67 Omega` |
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