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Alternating Current question

2024 · 30 Jan · Shift 1 · Q66
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Alternating Current question

2024 · 30 Jan · Shift 1 · Q66

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A series L.R circuit connected with an ac source E=(25sin⁡1000t)VE=(25 \sin 1000 t) VE=(25sin1000t)V has a power factor of 12\frac{1}{\sqrt{2}}2​1​. If the source of emf is changed to E=(20sin⁡2000t)V\mathrm{E}=(20 \sin 2000 \mathrm{t}) \mathrm{V}E=(20sin2000t)V, the new power factor of the circuit will be :
  1. A
    13\frac{1}{\sqrt{3}}3​1​
  2. B
    12\frac{1}{\sqrt{2}}2​1​
  3. C
    15\frac{1}{\sqrt{5}}5​1​
  4. D
    17\frac{1}{\sqrt{7}}7​1​
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