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Alternating Current question

2015 · Q123
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Alternating Current question

2015 · Q123

NEETPhysicsAlternating CurrentMCQ+4 / −1
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z'. the power drawn will be
  1. A
    P(RZ)P\left( {{R \over Z}} \right)P(ZR​)
  2. B
    P
  3. C
    P(RZ)2P{\left( {{R \over Z}} \right)^2}P(ZR​)2
  4. D
    PRZP\sqrt {{R \over Z}}PZR​​
View written solutionFree

Correct answer: C

For pure resistor circuit, power

P = Vrms2R{{V_{rms}^2} \over R}RVrms2​​

⇒\Rightarrow⇒ Vrms2{V_{rms}^2}Vrms2​ = PR

For L-R series circuit, power

P' = Vrms Irms cosϕ\phi ϕ

= Vrms×VrmsZ×RZ{V_{rms}} \times {{{V_{rms}}} \over Z} \times {R \over Z}Vrms​×ZVrms​​×ZR​

= Vrms2RZ2{{V_{rms}^2R} \over {{Z^2}}}Z2Vrms2​R​ = PR×RZ2{{PR \times R} \over {{Z^2}}}Z2PR×R​

∴\therefore∴ P' = P(RZ)2P{\left( {{R \over Z}} \right)^2}P(ZR​)2

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