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JEE Main 2014 (12 Apr, Shift 1) · Pure R, L and C Circuits

Pure R, L and C Circuits question from JEE Main 2014 (12 Apr, Shift 1)

Q63JEE Main 2014 · 12 APR · SHIFT 1Medium

Pure R, L and C Circuits

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

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Answer: (b)

  1. XL=ωL=500×0.02=10 ΩX_L = \omega L = 500 \times 0.02 = 10\ \Omega, so I0=10010=10I_0 = \frac{100}{10} = 10 A.
  2. In a pure inductor the current lags the voltage by π2\frac{\pi}{2}.
  3. I=10sin⁡(500t−π2)=−10cos⁡(500t)I = 10 \sin\left(500t - \frac{\pi}{2}\right) = -10 \cos(500t)
  4. So the answer is (b) −10cos⁡(500t)-10 \cos(500t).
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