MHT CET · Physics · Alternating Current
\(\mathrm{L}=2 \mathrm{H}, \mathrm{C}=5 \mathrm{mF}\) and \(\mathrm{R}=12 \Omega\) are connected in series to an a.c. generator of frequency 50 Hz . Then
- A at resonance, impedance of the circuit is zero.
- B at resonance, impedance of the circuit is \(12 \Omega\).
- C the resonant frequency of the circuit is \(1 / 2 \pi\).
- D the inductive reactance is less than the capacitive reactance.
Answer & Solution
Correct Answer
(B) at resonance, impedance of the circuit is \(12 \Omega\).
Step-by-step Solution
Detailed explanation
Impedance, \(\mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}\)
At resonance
\(\begin{array}{ll}
& \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\
\therefore \quad & \mathrm{Z}=\mathrm{R} \\
\therefore \quad & \mathrm{Z}=12 \Omega
\end{array}\)
At resonance
\(\begin{array}{ll}
& \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\
\therefore \quad & \mathrm{Z}=\mathrm{R} \\
\therefore \quad & \mathrm{Z}=12 \Omega
\end{array}\)
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