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When a resistance of 2 ohm is connected across terminals of a cell, the current is `0.5 A`. When the resistance is increased to 5 ohm, the current is `0.25 A` The e.m.f. of the cell isA. `1.0 V`B. `1.5 V`C. `2.0 V`D. `2.5 V` |
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Answer» Correct Answer - B (b) Since `i= ((E)/(R +r))`, we get `0.5 = (E)/(2 + r)` `0.25 = (E)/(5 + r)` Dividing (i) by (ii), we get `2 = (5 + r)/(2 + r) implies r = Omega` `:. 0.5 = (E)/(2 + 1) implies E = 1.5 V` |
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