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Five identical lamps , each of resistance `1100 Omega` are connected to `220 V` The reading of an ideal ammeter `A` is A. `(220)/(1100)xx 1A`B. `(220)/(1100)xx 2A`C. `(220)/(1100)xx 3A`D. `(220)/(1100)xx 5A` |
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Answer» Correct Answer - c In the given circuit the potential difference across each resistor is `220V` The currect through ammeter in the sum of currect flowing through the three resistance to parallel to the right of ammeter `:.` Current through ammeter `= 3 xx(220)/(11000)A` |
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