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Alternating Current question

2005 · Shift 0 · Q131
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Alternating Current question

2005 · Shift 0 · Q131

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
A circuit has a resistance of 12ohm12ohm12ohm and an impedance of 15ohm15ohm15ohm. The power factor of the circuit will be
  1. A
    0.40.40.4
  2. B
    0.80.80.8
  3. C
    0.1250.1250.125
  4. D
    1.251.251.25
View written solutionFree

Correct answer: B

  1. In an AC circuit, the power factor is

cos⁡ϕ=RZ\cos\phi = \frac{R}{Z}cosϕ=ZR​

where:

  • RRR = resistance
  • ZZZ = impedance
  1. Given:

R=12 Ω,Z=15 ΩR = 12\,\Omega, \qquad Z = 15\,\OmegaR=12Ω,Z=15Ω

So,

cos⁡ϕ=1215\cos\phi = \frac{12}{15}cosϕ=1512​

  1. Simplify:

1215=45=0.8\frac{12}{15} = \frac{4}{5} = 0.81512​=54​=0.8

  1. Therefore, the power factor of the circuit is

0.80.80.8

  1. Checking options:
  • A: 0.40.40.4 ❌
  • B: 0.80.80.8 ✅
  • C: 0.1250.1250.125 ❌
  • D: 1.251.251.25 ❌ (power factor cannot exceed 1)

Hence, the correct option is B.

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