A) 1/2
B) \[1/\sqrt{2}\]
C) 1
D) \[\sqrt{3}/2\]
Correct Answer: C
Solution :
Here, phase difference |
\[\tan \phi =\frac{{{X}_{L}}-{{X}_{C}}}{R}\] |
\[\tan \frac{\pi }{3}=\frac{{{X}_{L}}-{{X}_{C}}}{R}\] |
When, L is removed |
\[\sqrt{3}=\frac{{{X}_{C}}}{R}\] |
\[{{X}_{C}}=\sqrt{3}R\] |
When C is removed |
\[\tan \frac{\pi }{3}=\sqrt{3}=\frac{{{X}_{L}}}{R}\] |
\[{{X}_{L}}=R\sqrt{3}\] |
Hence, in resonant circuit |
\[\tan \phi =\frac{\sqrt{3}R=\sqrt{3}R}{R}=0\] |
\[\phi =0\] |
\[\therefore \] Power factor \[\cos \phi =1\] |
It is the condition of resonance therefore phase difference between voltage and current is zero and power factor is \[\cos \phi =1\]. |
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