Question

A resistance of 600 ohms, an inductor of $0.4H$ and a capacitor of $0.01\,\mu F$ are connected in series to an AC source of variable frequency .Find the frequency of the AC source for which current in the circuit is maximum. Also, calculate the bandwidth and quality factor for the circuit.

Answer 1

Hint: The current in an LRC circuit connected to an AC source is maximum when the frequency of the AC source is equal to the resonant frequency of the circuit. The bandwidth of a circuit is inversely proportional to its quality factor.

Formula used:
- The characteristic resonant frequency of a series LCR circuit is given as $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{1}{{LC}}} $ , where $L$ is the inductance of the inductor and $C$ is the capacitance of the capacitor in the circuit.
- Quality factor $Q = \dfrac{1}{R}\sqrt {\dfrac{L}{C}} $ , where $R$ is the resistance of the resistor in the circuit
- Bandwidth B.W. $ = \,\dfrac{f}{Q}$

Complete step by step solution:
We’ve been given:
$R = 600\,\Omega $
$L = 0.4H$
$C = 0.01\mu F$
The resonant frequency of a series RLC circuit is given as
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{1}{{LC}}} $
When the AC source has the frequency corresponding to the resonant frequency, the maximum current will flow in the circuit. Hence the frequency of the AC source can be calculated as:
$
  f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{1}{{0.4 \times 0.01 \times {{10}^{ - 6}}}}} \\
   = 0.159 \times \sqrt {250 \times {{10}^6}} \\
   = 2514.01\,Hz \\
 $
The quality factor $Q$ is defined as
$
  Q = \dfrac{1}{R}\sqrt {\dfrac{L}{C}} \\
   = \dfrac{1}{{600}}\sqrt {\dfrac{{0.4}}{{0.01 \times {{10}^{ - 6}}}}} \\
   = 10.54 \\
 $
The bandwidth (B.W.) of the circuit is defined as the ratio of the resonant frequency to the quality factor
$
  {\text{B}}{\text{.W}}{\text{.}} = \,\dfrac{f}{Q} \\
   = \dfrac{{2514.01}}{{10.54}} \\
   = 238.52\,Hz \\
 $

Additional Information:
The quality factor of an RLC circuit tells us how sharp the response of the circuit is to an AC power supply. It relates the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of the oscillation of the AC power supply.
The bandwidth tells us how selective the circuit is in its ability to reject frequencies as we move away from the resonant frequency where the current flow in the circuit is maximum. The higher the quality factor of a circuit, the smaller will be its bandwidth since the circuit will be very specific in picking out the frequencies close to its natural frequency.

Note:
The quality factor can alternatively be calculated as:
$Q = \dfrac{{{X_L}}}{R}$
where,${X_L} = 2\pi fL$ is also called the inductive impedance.
Hence,
$
  Q = \dfrac{{2\pi fL}}{R} \\
   = \dfrac{{2 \times \pi \times 2514.01 \times 0.4}}{{600}} \\
   = 10.54 \\
 $