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How many lamps each of `50W`and `100W`can be connected in parallel across a `120V` battery of internal resistance `10 Omega` so the each glows to full power ?A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - b Resistance of each lamp `R = V^(2)//P = (100)^(2)//50` `= 200 Omega` , current through each lamp `= P//V= 50//100 = 1//2A` If there n lamps in parallel to a battery of `120V` current each lamp glown to fall power then total current in n cell `= n xx 1//2 A ,` Total resistance of circuit `= 10 + 200//n` As, current `= ("emf")/("total resistance")` So `(n)/(2) = (120)/(10+200//n)` On solving we get , `n = 4` |
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