KCET · Physics · Alternating Current
An inductance of \(\left(\frac{200}{\pi}\right) \mathrm{mH}\), a capacitance of \(\left(\frac{10^{-3}}{\pi}\right) \mathrm{F}\) and a resistance of \(10 \Omega\) are connected in series with an AC source \(220 \mathrm{~V}, 50 \mathrm{~Hz}\). The phase angle of the circuit is
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
The phase angle \((\theta)\) between \(I\) and \(V\) is given by
\(\tan \theta=\frac{X_{L}-X_{C}}{R}\)
where, \(X_{L}=2 \pi f L\)
\(=2 \pi \times 50 \times\left[\frac{200}{\pi} \times 10^{-3}\right]=20 \Omega \)
\(X_{C} =\frac{1}{2 \pi f C}=\frac{1 \times \pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega\)
and \(R=10 \Omega\)
Substituting values of \(X_{L}, X_{C}\) and \(R\) is Eq. (i), we get
\(\tan \theta=\frac{20-10}{10}=1\) \(\Rightarrow \quad \tan \theta=\tan \frac{\pi}{4}\) \(\therefore \quad \theta=\frac{\pi}{4}\) The phase angle of the circuit is \(\frac{\pi}{4} .\)
\(\tan \theta=\frac{X_{L}-X_{C}}{R}\)
where, \(X_{L}=2 \pi f L\)
\(=2 \pi \times 50 \times\left[\frac{200}{\pi} \times 10^{-3}\right]=20 \Omega \)
\(X_{C} =\frac{1}{2 \pi f C}=\frac{1 \times \pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega\)
and \(R=10 \Omega\)
Substituting values of \(X_{L}, X_{C}\) and \(R\) is Eq. (i), we get
\(\tan \theta=\frac{20-10}{10}=1\) \(\Rightarrow \quad \tan \theta=\tan \frac{\pi}{4}\) \(\therefore \quad \theta=\frac{\pi}{4}\) The phase angle of the circuit is \(\frac{\pi}{4} .\)
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