Alternating Current

JEE Main level, single correct, medium

47 of 190

Question 47 of 190

A series ACAC circuit containing an inductor (20mH),\left(20 mH\right), a capacitor (120μF)\left(120\text{μ}F\right) and a resistor (60Ω)\left(60 \Omega \right) is driven by an ACAC source of 24V/50Hz.24 V/50 Hz. The energy dissipated in the circuit in 60s60 s is:

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Answer: (A) 5.17×102J5.17\times {10}^{2}J

(Power factor), Pf=cos⁡⁡ϕ=RZ{P}_{f}=\cos ⁡\text{ϕ}=\dfrac{R}{Z}
Where R,ZR,Z are the resistance and impedance respectively.
(Energy dissipated) E=Vrms2Z×cos⁡ϕ×tE=\dfrac{{V}_{rms}^{2}}{Z}\times \cos \text{ϕ}\times t
⇒E=Vrms2Z×RZ×t\Rightarrow E=\dfrac{{V}_{rms}^{2}}{Z}\times \dfrac{R}{Z}\times t
⇒E=Vrms2Z2×R×t\Rightarrow E=\dfrac{{V}_{rms}^{2}}{{Z}^{2}}\times R\times t
Vrms,t{V}_{rms},t are the RMS voltage and time respectively.We know that the Impedance of a LCR circuit is given by the formula,
Z2=R2+(ωL−1ωC)2⇒Z2=602+(2π×50×20×10−3−(12π×50×120×10−6))2{Z}^{2}={R}^{2}+{\left(\text{ω}L-\dfrac{1}{\text{ω}C}\right)}^{2} \Rightarrow {Z}^{2}={60}^{2}+{\left(2\text{π}\times 50\times 20\times {10}^{-3}-\left(\dfrac{1}{2\text{π}\times 50\times 120\times {10}^{-6}}\right)\right)}^{2}
Where ω,L,C\text{ω},L,C are the angular frequency, inductance and capacitance in the circuit respectively.
⇒Z2=602+(2π−100012π)2\Rightarrow {Z}^{2}={60}^{2}+{\left(2\text{π}-\dfrac{1000}{12\text{π}}\right)}^{2}

⇒Z2=4009.7\Rightarrow {Z}^{2}=4009.7

∴E=2424009.7×60×60J∴E=\dfrac{{24}^{2}}{4009.7}\times 60\times 60J
=517.14J= 517.14 J
=5.17×102J=5.17\times {10}^{2}J

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