JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 6)
A series L.R circuit connected with an ac source $$E=(25 \sin 1000 t) V$$ has a power factor of $$\frac{1}{\sqrt{2}}$$. If the source of emf is changed to $$\mathrm{E}=(20 \sin 2000 \mathrm{t}) \mathrm{V}$$, the new power factor of the circuit will be :
$$\frac{1}{\sqrt{3}}$$
$$\frac{1}{\sqrt{2}}$$
$$\frac{1}{\sqrt{5}}$$
$$\frac{1}{\sqrt{7}}$$
Explanation
$$\begin{aligned} & E=25 \sin (1000 t) \\\\ & \cos \theta=\frac{1}{\sqrt{2}} \end{aligned}$$
LR circuit
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$$\begin{aligned} & \text { Initially } \frac{R}{\omega_1 L}=\frac{1}{\tan \theta}=\frac{1}{\tan 45^{\circ}}=1 \\\\ & X_L=\omega_1 L \\\\ & \omega_2=2 \omega_1, \text { given } \\\\ & \tan \theta^{\prime}=\frac{\omega_2 L}{R}=\frac{2 \omega_1 L}{R} \\\\ & \tan \theta^{\prime}=2 \\\\ & \cos \theta^{\prime}=\frac{1}{\sqrt{5}} \end{aligned}$$
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