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JEE Main 2020 (6 Sep, Shift 2) · Power, Power Factor & Wattless Current

Power, Power Factor & Wattless Current question from JEE Main 2020 (6 Sep, Shift 2)

Q62JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Power, Power Factor & Wattless Current

In a series LR circuit, power of 400W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as (n3π)μ\left(\frac{n}{3\pi}\right)\muF, then value of n is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 400

  1. P=VIcos⁡ϕ⇒I=400250×0.8=2P = V I\cos\phi \Rightarrow I = \frac{400}{250\times 0.8} = 2 A
  2. Z=250/2=125 ΩZ = 250/2 = 125\ \Omega, R=Zcos⁡ϕ=100 ΩR = Z\cos\phi = 100\ \Omega, XL=1252−1002=75 ΩX_L = \sqrt{125^2-100^2} = 75\ \Omega
  3. Unity power factor needs XC=75 ΩX_C = 75\ \Omega: C=1100π×75=17500πC = \frac{1}{100\pi\times 75} = \frac{1}{7500\pi} F
  4. C=1067500π μC = \frac{10^6}{7500\pi}\ \muF =4003π μ= \frac{400}{3\pi}\ \muF, so n=400n = 400.
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