MHT CET · Physics · Alternating Current
The coil of inductance \(0.25 \mathrm{mH}\) has a reactance of \(330 \Omega\), when connected to an a.c. source. The frequency of a.c. source is (Take \(\left.\pi=\frac{22}{7}\right)\)
- A \(210 \mathrm{kHz}\)
- B \(105 \mathrm{kHz}\)
- C \(420 \mathrm{kHz}\)
- D \(330 \mathrm{kHz}\)
Answer & Solution
Correct Answer
(A) \(210 \mathrm{kHz}\)
Step-by-step Solution
Detailed explanation
We know inductive reactance is given by \(X_L=2 \pi f L\), where \(f\) is the frequency of the source and \(L\) the inductance of the coil.
\(f=\frac{X_L}{2 \pi L}\)
Given, \(L=0.25 \mathrm{mH}\) and \(C=330 \Omega\),
\(f=\frac{330 \Omega}{2 \pi(0.25 \mathrm{mH})}=210 \mathrm{kHz}\)
\(f=\frac{X_L}{2 \pi L}\)
Given, \(L=0.25 \mathrm{mH}\) and \(C=330 \Omega\),
\(f=\frac{330 \Omega}{2 \pi(0.25 \mathrm{mH})}=210 \mathrm{kHz}\)
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