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JEE Main & Advanced Sample Paper JEE Main Sample Paper-2

  • question_answer
    An ideal solenoid stores 32 J of magnetic field energy and when a current of 4 Amp is passed through it, the rate of energy dissipation becomes 320 watt in the form of heat. The time constant of the circuit is (assume ideal battery connected)

    A)  0.4 s                     

    B)  0.314 s

    C)  0.2 s                                     

    D)  0.625 s

    Correct Answer: C

    Solution :

    Energy stored in inductor \[U=\frac{1}{2}L{{i}^{2}}\] \[\Rightarrow \]               \[32=\frac{1}{2}L{{(4)}^{2}}\] \[\Rightarrow \]               \[L=4\,H\] Rate of heat loss \[A={{i}^{2}}R\] \[320={{(4)}^{2}}R\] \[\Rightarrow \,\,R=20\,\Omega \] Time constant \[t=\frac{L}{R}=\frac{4}{20}=\frac{1}{5}=0.2\,s\]


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