A series RLC circuit has R = 5 W, L = 0.2 H and C = 50 µF. The applied voltage is 200 V. Find (i) resonant frequency (ii) Q-factor (iii) bandwidth (iv) upper and lower half-power frequencies (v) current at resonance (vi) current at half-power points (vii) voltage across inductance at resonance.
The correct answer and explanation is :
Given the following values for the RLC circuit:
- ( R = 5 \, \Omega )
- ( L = 0.2 \, \text{H} )
- ( C = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} )
- ( V = 200 \, \text{V} ) (applied voltage)
(i) Resonant Frequency
The resonant frequency ( f_0 ) for an RLC circuit is given by the formula:
[
f_0 = \frac{1}{2\pi \sqrt{LC}}
]
Substituting the values of ( L ) and ( C ):
[
f_0 = \frac{1}{2\pi \sqrt{0.2 \times 50 \times 10^{-6}}}
]
[
f_0 \approx \frac{1}{2\pi \sqrt{0.00001}} = \frac{1}{2\pi \times 0.00316} \approx 50.5 \, \text{Hz}
]
So, the resonant frequency is approximately 50.5 Hz.
(ii) Q-factor
The quality factor (Q-factor) of the circuit is given by:
[
Q = \frac{f_0}{\Delta f}
]
Where ( \Delta f ) is the bandwidth. However, ( Q ) can also be expressed as:
[
Q = \frac{1}{R} \sqrt{\frac{L}{C}}
]
Substituting the known values:
[
Q = \frac{1}{5} \sqrt{\frac{0.2}{50 \times 10^{-6}}} \approx \frac{1}{5} \sqrt{4000} \approx \frac{1}{5} \times 63.25 = 12.65
]
So, the Q-factor is approximately 12.65.
(iii) Bandwidth
The bandwidth ( \Delta f ) of the circuit is related to the Q-factor by:
[
\Delta f = \frac{f_0}{Q}
]
Substituting the values:
[
\Delta f = \frac{50.5}{12.65} \approx 4.0 \, \text{Hz}
]
So, the bandwidth is approximately 4.0 Hz.
(iv) Upper and Lower Half-Power Frequencies
The upper and lower half-power frequencies are given by:
[
f_{\text{upper}} = f_0 + \frac{\Delta f}{2}, \quad f_{\text{lower}} = f_0 – \frac{\Delta f}{2}
]
Substituting the values:
[
f_{\text{upper}} = 50.5 + \frac{4.0}{2} = 52.5 \, \text{Hz}
]
[
f_{\text{lower}} = 50.5 – \frac{4.0}{2} = 48.5 \, \text{Hz}
]
So, the upper half-power frequency is approximately 52.5 Hz and the lower half-power frequency is approximately 48.5 Hz.
(v) Current at Resonance
The current at resonance ( I_0 ) is given by:
[
I_0 = \frac{V}{R}
]
Substituting the known values:
[
I_0 = \frac{200}{5} = 40 \, \text{A}
]
So, the current at resonance is 40 A.
(vi) Current at Half-Power Points
At the half-power points, the voltage and current drop by ( \frac{1}{\sqrt{2}} ) of their peak values. The current at half-power frequencies can be calculated as:
[
I_{\text{half}} = \frac{I_0}{\sqrt{2}} \approx \frac{40}{\sqrt{2}} \approx 28.3 \, \text{A}
]
So, the current at half-power points is approximately 28.3 A.
(vii) Voltage Across Inductance at Resonance
At resonance, the voltage across the inductor is equal to the current multiplied by the inductive reactance ( X_L ):
[
X_L = 2\pi f_0 L
]
Substituting the known values:
[
X_L = 2\pi \times 50.5 \times 0.2 = 63.5 \, \Omega
]
The voltage across the inductance is:
[
V_L = I_0 \times X_L = 40 \times 63.5 = 2540 \, \text{V}
]
So, the voltage across the inductance at resonance is 2540 V.
Conclusion:
- Resonant frequency: 50.5 Hz
- Q-factor: 12.65
- Bandwidth: 4.0 Hz
- Upper half-power frequency: 52.5 Hz
- Lower half-power frequency: 48.5 Hz
- Current at resonance: 40 A
- Current at half-power points: 28.3 A
- Voltage across inductance at resonance: 2540 V
This analysis provides a comprehensive view of the circuit’s behavior, including resonance characteristics and the impact on current and voltage at various frequencies.