JEE Mains · Physics · STD 12 -7. Alternating current
A sinusoidal voltage \(V(t)=210\,\sin 3000 t\) volt is applied to a series \(LCR\) circuit in which \(L =10\) \(mH , C =25 \mu F\) and \(R =100 \Omega\). The phase difference \((\Phi)\) between the applied voltage and resultant current will be
- A \(\tan ^{-1}(0.17)\)
- B \(\tan ^{-1}(9.46)\)
- C \(\tan ^{-1}(0.30)\)
- D \(\tan ^{-1}(13.33)\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}(0.17)\)
Step-by-step Solution
Detailed explanation
\(X _{ L }=10^{-2} \times 3000=30 \Omega\) \(X _{ C }=\frac{1}{3000 \times 25 \times 10^{-6}}=\frac{40}{3} \Omega\) \(X = X _{ L }- X _{ C }\) \(=30-\frac{40}{3}=\frac{50}{3}\) \(\tan \delta=\frac{ X }{ R }=\frac{50}{3 \times 100}=\frac{1}{6}\)…
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