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GUJCET · Physics · Alternating Current
A pure inductor of \(\mathbf{2 5 . 4 8 ~ m H}\) and a pure resistor of \(8 \Omega\) are connected in series with an AC source of frequency 50 Hz . The phase difference between current (I) and voltage (V) in this circuit is ______________ .
- A \(45^{\circ}\)
- B \(30^{\circ}\)
- C \(60^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(A) \(45^{\circ}\)
Step-by-step Solution
Detailed explanation
(A) \(45^{\circ}\)
\(\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \nu \mathrm{~L}\)
\(\mathrm{X}_{\mathrm{L}}=2 \times 3.14 \times 50 \times 25.48 \times\) \(10^{-3}\)
\(\mathrm{X}_{\mathrm{L}}=8000.72 \times 10^{-3}\)
\(\mathrm{X}_{\mathrm{L}}=8 \Omega\)
phase difference \(\tan \delta=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)
\(
\begin{aligned}
\tan \delta & =\frac{8}{8} \\
\tan \delta & =1 \\
\delta & =45^{\circ}
\end{aligned}
\)
\(\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \nu \mathrm{~L}\)
\(\mathrm{X}_{\mathrm{L}}=2 \times 3.14 \times 50 \times 25.48 \times\) \(10^{-3}\)
\(\mathrm{X}_{\mathrm{L}}=8000.72 \times 10^{-3}\)
\(\mathrm{X}_{\mathrm{L}}=8 \Omega\)
phase difference \(\tan \delta=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)
\(
\begin{aligned}
\tan \delta & =\frac{8}{8} \\
\tan \delta & =1 \\
\delta & =45^{\circ}
\end{aligned}
\)
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