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Alternating Current — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current4 Marks
Question
When the frequency of ac supply is such that the inductive reactance and capacitive reactance become equal, the impedance of the series $\text{LCR}$ circuit is equal to the ohmic resistance in the circuit. Such a series $\text{LCR}$ circuit is known as resonant series $\text{LCR}$ circuit and the frequency of the ac supply is known as resonant frequency. Resonance phenomenon is exhibited by a circuit only if both Land Care present in the circuit. We cannot have resonance in a $\text{RL}$ or $\text{RC}$ circuit.
A series $\text{LCR}$ circuit with $L = 0.12H, C = 480\ nF, \text{R} = 23\Omega$ is connected to a $230V$ variable frequency supply.
  1. Find the value of source frequency for which current amplitude is maximum.
  1. $222.32\ Hz$
  2. $550.52\ Hz$
  3. $663.48\ Hz$
  4. $770\ Hz$
  1. The value of maximum current is:
  1. $14.14A$
  2. $22.52A$
  3. $50.25A$
  4. $47.41A$
  1. The value of maximum power is:
  1. $2200W$
  2. $2299.3W$
  3. $5500W$
  4. $4700W$
  1. What is the $Q-$factor of the given circuit?
  1. $25$
  2. $42.21$
  3. $35.42$
  4. $21.74$
  1. At resonance which of the following physical quantity is maximum?
  1. Impedance
  2. Current
  3. Both $(a)$ and $(b)$
  4. Neither $(a)$ nor $(b)$
✓

Answer

  1. $(c)\ 663.48\ Hz$
Here, $L = 0.12H, e = 480\ nF = 480 \times 10^{-9}F$
$\text{R} = 23\Omega, V = 230V$
$\text{V}_0=\sqrt{2}\times230=325.22\text{V}$
$\text{I}_0=\frac{\text{V}_0}{\sqrt{\text{R}^2+\Big(\omega\text{L}-\frac{1}{\omega\text{C}}\Big)^2}}$
At resonance, $\omega\text{L}-\frac{1}{\omega\text{C}}=0$
$\omega=\frac{1}{\sqrt{\text{LC}}}=\frac{1}{\sqrt{0.12\times480\times10^9}}=4166.67\ \text{rad}\ \text{s}^{-1}$
$\upsilon_\text{R}=\frac{4166.67}{2\times3.14}=663.48\text{H}_\text{z}$
  1. $(a)\ 14.14A$
Current, $\text{I}_0=\frac{\text{V}_0}{\text{R}}=\frac{325.22}{23}=14.14\text{A}$
  1. $(b)\ 2299.3W$
Maximum power, $\text{P}_\text{max}=\frac{1}{2}(\text{I}_0)^2\text{R}$
$=\frac{1}{2}\times(14.14)^2\times23=2299.3\text{W}$
  1. $(d)\ 21.74$
Quality factor, $\text{Q}=\frac{\text{X}_\text{R}}{\text{R}}=\frac{\omega_\text{r}\text{L}}{\text{R}}$
$=\frac{4166.67\times0.12}{23}=21.74$
  1. $(b)$ Current

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