1}})=10\,A\]; Final current \[({{I}_{2}})=0\]; TIME (t) = 0.5 sec and INDUCED e.m.f. \[(\varepsilon )=220\] V. Induced e.m.f. \[(\varepsilon )\] \[=-L\frac{dI}{DT}=-L\frac{({{I}_{2}}-{{I}_{1}})}{t}=-L\frac{(0-10)}{0.5}=20L\] Or, \[L=\frac{220}{20}=11H\] [where L = Self inductance of coil]

"> 1}})=10\,A\]; Final current \[({{I}_{2}})=0\]; TIME (t) = 0.5 sec and INDUCED e.m.f. \[(\varepsilon )=220\] V. Induced e.m.f. \[(\varepsilon )\] \[=-L\frac{dI}{DT}=-L\frac{({{I}_{2}}-{{I}_{1}})}{t}=-L\frac{(0-10)}{0.5}=20L\] Or, \[L=\frac{220}{20}=11H\] [where L = Self inductance of coil]

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A varying current in a coil changes from 10A to zero in 0.5 sec. If the average e.m.f induced in the coil is 220V, the self-inductance of the coil is

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 1 year ago

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[c] Initial current \[({{I}_{1}})=10\,A\]; Final current \[({{I}_{2}})=0\]; TIME (t) = 0.5 sec and INDUCED e.m.f. \[(\varepsilon )=220\] V. Induced e.m.f. \[(\varepsilon )\] \[=-L\frac{dI}{DT}=-L\frac{({{I}_{2}}-{{I}_{1}})}{t}=-L\frac{(0-10)}{0.5}=20L\] Or, \[L=\frac{220}{20}=11H\] [where L = Self inductance of coil]

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Posted on 11 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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