A) 0.5
B) 1.0
C) -1.0
D) zero
Correct Answer: B
Solution :
[b] When L is removed, |
\[\tan \phi =\frac{\left| {{X}_{C}} \right|}{R}\Rightarrow \tan \frac{\pi }{3}=\frac{{{X}_{C}}}{R}\] (i) |
When C is removed, |
\[\tan \phi =\frac{\left| {{X}_{L}} \right|}{R}\Rightarrow \tan \frac{\pi }{3}=\frac{{{X}_{L}}}{R}\] (ii) |
From (i) and (ii), \[{{X}_{L}}={{X}_{C}}\] |
Since, \[{{X}_{L}}={{X}_{C}}\], the circuit is in resonance. |
\[Z=R\] |
Power factor \[=\cos \phi =\frac{R}{Z}=1\] |
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