FRAC{\pi }{2}\]. If \[v(t)={{v}_{0}}\,\sin \,\omega t\] then \[I={{I}_{0}}\,\sin \,\left( \omega t-\frac{\pi }{2} \right)\] Now, GIVEN v(t) = 100 sin (500 t) and    \[{{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\]   \[[\because \,L=0.02\,H]\] \[{{I}_{0}}=10\,\sin \,\left( 500t-\frac{\pi }{2} \right)=-10\,\cos \,(500\,t)\]

"> FRAC{\pi }{2}\]. If \[v(t)={{v}_{0}}\,\sin \,\omega t\] then \[I={{I}_{0}}\,\sin \,\left( \omega t-\frac{\pi }{2} \right)\] Now, GIVEN v(t) = 100 sin (500 t) and    \[{{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\]   \[[\because \,L=0.02\,H]\] \[{{I}_{0}}=10\,\sin \,\left( 500t-\frac{\pi }{2} \right)=-10\,\cos \,(500\,t)\]

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A sinusoidal voltage V(t) = 100 sin (500t) is applied across a pure inductance of L=0.02H. The current through the coil is:

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 9 months ago

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[b] In a pure inductive circuit current always lags behind the emf by\[\FRAC{\pi }{2}\]. If \[v(t)={{v}_{0}}\,\sin \,\omega t\] then \[I={{I}_{0}}\,\sin \,\left( \omega t-\frac{\pi }{2} \right)\] Now, GIVEN v(t) = 100 sin (500 t) and    \[{{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\]   \[[\because \,L=0.02\,H]\] \[{{I}_{0}}=10\,\sin \,\left( 500t-\frac{\pi }{2} \right)=-10\,\cos \,(500\,t)\]

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Posted on 07 Sep 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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