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Two resistances are joined in parallel whose equivolent resistance is `3/5Omega`. One of the resistance wire is broken and the effective resistance becomes `3Omega`. The resistance (in ohms) of the wire that got broken wasA. `4/3`B. 2C. `6/5`D. `3/4` |
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Answer» Correct Answer - D Resistance of the wire `R_(P)=(R_(1)R_(2))/(R_(1)+R_(2))=(3)/(5)` and `R_(1)=3 Omega`, then `therefore (3xxR_(2))/(3+R_(2))=(3)/(5)rArr 15 R_(2)=9+3 R_(2)rArr 12 R_(2)=9` `therefore R_(2)=(3)/(4)Omega` |
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