JEE Mains · Physics · STD 12 -7. Alternating current
A series \(LCR\) circuit is connected to an \(AC\) source of \(220\,V , 50\,Hz\). The circuit contains a resistance \(R =80\,\Omega\), an inductor of inductive reactance \(X _{ L }=70\,\Omega\), and a capacitor of capacitive reactance \(X _{ C }=130\,\Omega\). The power factor of circuit is \(\frac{ X }{10}\). The value of \(x\) is:
- A \(4\)
- B \(8\)
- C \(6\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\(\cos \phi=\frac{ R }{ Z }=\frac{ R }{\sqrt{ R ^2+\left( X _{ C }- X _{ L }\right)^2}}\) \(\cos \phi=\frac{80}{\sqrt{(80)^2+(60)^2}}\) \(\cos \phi=\frac{80}{100} \Rightarrow \frac{8}{10}\) So, \(x=8\)
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