A) \[\frac{\sqrt{5}R}{2},\,{{\tan }^{-1}}(2)\]
B) \[\frac{\sqrt{5}R}{2},{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
C) \[\sqrt{5}{{X}_{C}},{{\tan }^{-1}}(2)\]
D) \[\sqrt{5}R,\,{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
Correct Answer: B
Solution :
\[{{X}_{L}}=R,\ \ {{X}_{C}}=R/2\] \[\therefore \tan \varphi =\frac{{{X}_{L}}-{{X}_{C}}}{R}=\frac{R-\frac{R}{2}}{R}=\frac{1}{2}\] \[\Rightarrow \varphi ={{\tan }^{-1}}(1/2)\] Also\[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}=\sqrt{{{R}^{2}}+\frac{{{R}^{2}}}{4}}=\frac{\sqrt{5}}{2}R\]
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