JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    In an a.c. circuit, the instantaneous e.m.f. and current are given by\[\text{e=100 sin 30 t}\]\[\text{i=20 sin }\left( 30t-\frac{\pi }{4} \right)\]. In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively: [JEE Main Online 08-04-2018]

    A)  \[\frac{50}{\sqrt{2}},0\]                   

    B)  \[50,\,\,0\]

    C)  \[50,\,\,10\]                 

    D)  \[\frac{1000}{\sqrt{2}},\,\,10\]

    Correct Answer: D

    Solution :

    \[e=100\sin (30t)\] \[i=20\sin \left( 30t-\frac{\pi }{4} \right)\] \[{{e}_{rms}}=\frac{100}{\sqrt{2}}volt,\]               \[{{i}_{rms}}=\frac{20}{\sqrt{2}}amp\] Power factor \[=\cos \phi =cos\left( -\frac{\pi }{4} \right)=\frac{1}{\sqrt{2}}\] \[<P>={{e}_{rms}}{{i}_{rms}}\cos \phi \] \[=\frac{100}{\sqrt{2}}\times \frac{20}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\] \[=\frac{1000}{\sqrt{2}}watt\] Wattles current\[={{i}_{rms}}\sin \phi \] \[=\frac{20}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=10amp\]


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