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Question

Which of the following combinations should be used for better tuning of an LCR circuit used for communication?

The correct answer is

R = 15Ω, L = 3.5H, C = 30 μF

Understanding LCR Circuit Tuning for Communication

LCR circuits are commonly used in communication systems, particularly in radio receivers, as tuning circuits. The ability of the circuit to select a specific frequency from a range of frequencies is called selectivity. Better tuning means higher selectivity, allowing the circuit to sharply respond to the desired frequency while rejecting nearby frequencies.

The selectivity of an LCR circuit at resonance is related to its bandwidth. A narrower bandwidth means higher selectivity and thus better tuning. The bandwidth ($\Delta \omega$) of a series LCR circuit at resonance is given by the formula:

$\Delta \omega = \frac{R}{L}$

where R is the resistance and L is the inductance in the circuit.

For better tuning (narrower bandwidth), the ratio $\frac{R}{L}$ should be as small as possible. We need to examine the given options and calculate the $\frac{R}{L}$ ratio for each combination of R and L.

Analyzing R and L Values for Each Option

Let's calculate the ratio $\frac{R}{L}$ for each set of parameters provided in the options:

Option R ($\Omega$) L (H) $\frac{R}{L}$ (s$^{-1}$) Bandwidth
1 25 1.5 $\frac{25}{1.5} \approx 16.67$ Wider
2 25 2.0 $\frac{25}{2.0} = 12.5$ Wider
3 15 3.5 $\frac{15}{3.5} \approx 4.29$ Narrowest
4 15 1.0 $\frac{15}{1.0} = 15.0$ Wider

Determining the Best LCR Combination for Tuning

Comparing the calculated $\frac{R}{L}$ ratios, we see that the ratio is smallest for Option 3 ($15/3.5 \approx 4.29$). A smaller $\frac{R}{L}$ ratio results in a narrower bandwidth, which in turn provides better selectivity and tuning for the LCR circuit in communication applications.

While the capacitance (C) is important for determining the resonant frequency ($\omega_0 = \frac{1}{\sqrt{LC}}$), it does not directly affect the bandwidth or selectivity ($\Delta \omega$) based on the $\frac{R}{L}$ formula alone. However, the quality factor (Q) is another measure of selectivity, defined as $Q = \frac{\omega_0 L}{R} = \frac{1}{R} \sqrt{\frac{L}{C}}$. A higher Q factor indicates better selectivity. Since $\Delta \omega = \frac{\omega_0}{Q}$, maximizing Q is equivalent to minimizing $\frac{R}{L}$ for a given resonant frequency or minimizing $\frac{R}{L}$ generally for the narrowest bandwidth.

Based on the analysis of the $\frac{R}{L}$ ratio, the combination with R = 15$\Omega$ and L = 3.5H provides the minimum ratio, leading to the best tuning.

Revision Table: LCR Circuit Key Concepts

Concept Definition/Formula Effect on Tuning
Resonant Frequency ($\omega_0$) $\omega_0 = \frac{1}{\sqrt{LC}}$ Frequency at which the circuit impedance is minimum (series) or maximum (parallel). Determines the frequency tuned.
Bandwidth ($\Delta \omega$) $\Delta \omega = \frac{R}{L}$ (for series LCR) Range of frequencies around resonance where the circuit response is significant. Smaller bandwidth means better tuning/selectivity.
Quality Factor (Q) $Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$ Measure of the sharpness of resonance. Higher Q means better tuning/selectivity.
Selectivity Ability to distinguish between the resonant frequency and nearby frequencies. High selectivity means narrow bandwidth and high Q.

Additional Information: LCR Circuits and Selectivity

LCR circuits exhibit resonance when the inductive reactance ($X_L = \omega L$) equals the capacitive reactance ($X_C = \frac{1}{\omega C}$). This occurs at the resonant frequency $\omega_0$. In a series LCR circuit, the impedance is $Z = R + j(\omega L - \frac{1}{\omega C})$. At resonance, $Z = R$, which is the minimum impedance, leading to maximum current for a given voltage.

The quality factor (Q) is a dimensionless parameter that describes how underdamped an oscillator or resonator is. For an LCR circuit, a higher Q means the oscillations die out more slowly, and the resonance peak is sharper. This sharpness of the resonance peak directly relates to the circuit's selectivity. A sharp peak means the circuit responds strongly only to frequencies very close to the resonant frequency.

Factors affecting selectivity in an LCR circuit:

  • Resistance (R): Higher resistance leads to a wider bandwidth ($\Delta \omega = R/L$) and a lower quality factor ($Q = \omega_0 L/R$). Lower resistance improves selectivity.
  • Inductance (L): Higher inductance leads to a narrower bandwidth ($\Delta \omega = R/L$) and a higher quality factor ($Q = \omega_0 L/R$). Higher inductance improves selectivity (assuming R is constant or doesn't increase proportionally).
  • Capacitance (C): Affects the resonant frequency and the quality factor ($Q = \frac{1}{R}\sqrt{\frac{L}{C}}$). For a fixed resonant frequency, increasing L requires decreasing C, which tends to increase Q (improving selectivity), while decreasing L requires increasing C, which tends to decrease Q (reducing selectivity).

In summary, for better tuning and higher selectivity in an LCR circuit used for communication, we aim for a low resistance R and a high inductance L, minimizing the R/L ratio and maximizing the Q factor at the desired resonant frequency.

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Important Questions from Communication Systems

  1. The wavelength of radiation emitted when He+ makes a transition from the state n = 3 to the state n = 2 will be:

    (Take Rydberg constant R = 1.097 × 10⁷ m⁻¹)

  2. Match List - I with List - II 

    List-IList-II
    (A) Range(I) Range of frequencies over which communication system works
    (B) Band width(II) The largest distance between transmitter and receiver
    (C) Attenuation(III) Loss of strength of a signal during propagation
    (D) Transducer(IV) A device that receives an input in electrical form or provides an output in electrical form

    Choose the correct answer from the options given below:

  3. A carrier wave of peak voltage 14 V is used to transmit a message. What should be the peak voltage of the modulating signal in order to have a modulation index of 70%?

  4. Match List - I with List - II

    List-IList-II
    (A) Range(I) Range of frequencies over which communication system works
    (B) Band width(II) The largest distance between transmitter and receiver
    (C) Attenuation(III) Loss of strength of a signal during propagation
    (D) Transducer(IV) A device that receives an input in electrical form or provides an output in electrical form

    Choose the correct answer from the options given below:

  5. A carrier wave of peak voltage 14 V is used to transmit a message. What should be the peak voltage of the modulating signal in order to have a modulation index of 70%?

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