Alternating Current

JEE Main level, numerical answer, medium

124 of 190

Question 124 of 190

In a series LRLR circuit, power of 400W400W is dissipated from a source of 250V,50Hz250V,50Hz. The power factor of the circuit is 0.80.8. In order to bring the power factor to unity, a capacitor of value CC is added in series to the LL and RR. Taking the value of CC as (n3π)μF\left(\dfrac{n}{3\pi }\right)\mu F, then value of nn is

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Answer: 400

P=VmP={V}_{m}. in cosϕcos\phi
400=250×1m×0.8400=250\times 1m\times 0.8
irms=2A{i}_{rms}=2A
(1m)2.R=P{\left(1m\right)}^{2}.R=P
4×R=4004\times R=400
⇒\Rightarrow R=100ΩR=100\Omega.
cosϕ=RR2+XL2cos\phi =\dfrac{R}{\sqrt{{R}^{2}+{X}_{L}^{2}}}
1002+XL2=(1000.8)2{100}^{2}+{X}_{L}^{2}={\left(\dfrac{100}{0.8}\right)}^{2}
1002+XL2=(1000.8)2{100}^{2}+{X}_{L}^{2}={\left(\dfrac{100}{0.8}\right)}^{2}
XL=75Ω{X}_{L}=75\Omega
Power factor is unity
XC=XL=75{X}_{C}={X}_{L}=75
1ω=75\dfrac{1}{\omega }=75
⇒\Rightarrow C=175×2H×50=17500πFC=\dfrac{1}{75\times 2H\times 50}=\dfrac{1}{7500\pi }F
3π×25003\pi \times 2500
=13π×4×102mF=\dfrac{1}{3\pi }\times 4\times {10}^{2} mF
=4003πμF=\dfrac{400}{3\pi }\mu F
N=400N=400

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