A) \[P{{\left( \frac{R}{Z} \right)}^{2}}\]
B) \[P\sqrt{\frac{R}{Z}}\]
C) \[P\left( \frac{R}{Z} \right)\]
D) P
Correct Answer: A
Solution :
When a resistor is connected to an AC source. The power drawn will be \[p={{V}_{rms}}/{{I}_{rms}}={{V}_{rms}}.\frac{{{V}_{rms}}}{R}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,V_{rms}^{2}=PR\] When an inductor is connected in series with the resistor, then the power drawn will be \[P'={{V}_{rms}}.{{I}_{rms}}\cos \phi \] where, \[\phi \]= phase difference \[\therefore \] \[P'=\frac{V_{rms}^{2}}{R}.\frac{{{R}^{2}}}{{{Z}^{2}}}=p.R.\frac{R}{{{Z}^{2}}}\] \[\Rightarrow \,\,\,\,\,p'=\frac{p.{{R}^{2}}}{{{Z}^{2}}}=P{{\left( \frac{R}{Z} \right)}^{2}}\]
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