A Series L R Circuit... |JEE Main 2024

JEE-Main_2024

30.01.24_(S1)

Question

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

Select the correct option:

1/√2

1/√3

1/√5

1/√7

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

E=25sin⁡(1000t)
cos⁡θ=1/√2
LR circuit

$\begin{aligned} & \text { Initially } \frac{R}{\omega_1 L}=\frac{1}{\tan \theta}=\frac{1}{\tan 45^{\circ}}=1 \\ & X_L=\omega_1 L \\ & \omega_2=2 \omega_1, \text { given } \\ & \tan \theta^{\prime}=\frac{\omega_2 L}{R}=\frac{2 \omega_1 L}{R} \\ & \tan \theta^{\prime}=2 \\ & \cos \theta^{\prime}=\frac{1}{\sqrt{5}}\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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