Verified Solution • 2025–26 Session moderate

QJEEM24Ph9388: JEE Main 2024 Physics - Class 12 JEEM24Ph9388 Solution: A series L.R circuit connected with an ac source has a power…

A series L.R circuit connected with an ac source E=(25sin⁡1000t)VE=(25 \sin 1000 t) V has a power factor of 12\frac{1}{\sqrt{2}} . If the source of emf is changed to E=(20sin⁡2000t)V\mathrm{E}=(20 \sin 2000 \mathrm{t}) \mathrm{V} , the new power factor of the circuit will be :

Multiple Choice Options

  • A 13\frac{1}{\sqrt{3}}
  • B 12\frac{1}{\sqrt{2}}
  • C 15\frac{1}{\sqrt{5}}
  • D 17\frac{1}{\sqrt{7}}

Step-by-Step Verified Solution

amp;E=25sin⁡(1000t)amp;cos⁡θ=12\begin{aligned}& E=25 \sin (1000 t) \\\\& \cos \theta=\frac{1}{\sqrt{2}}\end{aligned}

LR circuit

JEE Main 2024 (Online) 30th January Morning Shift Physics - Alternating Current Question 20 English Explanation

amp; Initially Rω1L=1tan⁡θ=1tan⁡45∘=1amp;XL=ω1Lamp;ω2=2ω1, given amp;tan⁡θ′=ω2LR=2ω1LRamp;tan⁡θ′=2amp;cos⁡θ′=15\begin{aligned}& \text { Initially } \frac{R}{\omega_1 L}=\frac{1}{\tan \theta}=\frac{1}{\tan 45^{\circ}}=1 \\\\& X_L=\omega_1 L \\\\& \omega_2=2 \omega_1, \text { given } \\\\& \tan \theta^{\prime}=\frac{\omega_2 L}{R}=\frac{2 \omega_1 L}{R} \\\\& \tan \theta^{\prime}=2 \\\\& \cos \theta^{\prime}=\frac{1}{\sqrt{5}}\end{aligned}

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