Understanding the Series RLC Circuit

A series RLC circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series. When this circuit is driven by an AC voltage source, its behavior depends on the frequency of the source. Important characteristics of a series RLC circuit include its resonant frequency and bandwidth.

• Resonant Frequency (\(\omega_0\)): This is the frequency at which the inductive reactance (\(X_L = \omega L\)) equals the capacitive reactance (\(X_C = \frac{1}{\omega C}\)). At resonance, the impedance of the circuit is purely resistive and is equal to R. The current in the circuit is maximum at this frequency. The formula for resonant frequency is \(\omega_0 = \frac{1}{\sqrt{LC}}\) (in rad/sec).
• Bandwidth (\(\Delta\omega\)): This is the range of frequencies over which the circuit's power output is at least half of the power output at resonance. For a series RLC circuit, the bandwidth is given by \(\Delta\omega = \frac{R}{L}\) (in rad/sec).

In this problem, we are given the bandwidth, resonant frequency, inductance, and the source voltage. We need to find the value of the resistance (R) and the voltage across the capacitor (V_C) at the resonant frequency.

Calculating Resistance (R) using Bandwidth

The bandwidth (\(\Delta\omega\)) of a series RLC circuit is related to the resistance (R) and inductance (L) by the formula:

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

We are given:

• Bandwidth, \(\Delta\omega = 300 \text{ rad/sec}\)
• Inductance, \(L = 0.1 \text{ H}\)

We can rearrange the formula to solve for R:

$$ R = \Delta\omega \times L $$

Substituting the given values:

$$ R = 300 \text{ rad/sec} \times 0.1 \text{ H} $$

$$ R = 30 \text{ } \Omega $$

So, the value of the resistance is 30 \(\Omega\).

Calculating Capacitance (C) (Intermediate Step)

Although we don't directly need C to find the voltage across the capacitor using a specific approach (as shown below), we can calculate it using the resonant frequency formula:

$$ \omega_0 = \frac{1}{\sqrt{LC}} $$

Squaring both sides gives:

$$ \omega_0^2 = \frac{1}{LC} $$

Rearranging to solve for C:

$$ C = \frac{1}{\omega_0^2 L} $$

We are given the resonant frequency, \(\omega_0 = 3000 \text{ rad/sec}\), and inductance, \(L = 0.1 \text{ H}\).

$$ C = \frac{1}{(3000 \text{ rad/sec})^2 \times 0.1 \text{ H}} = \frac{1}{(9 \times 10^6) \times 0.1} = \frac{1}{9 \times 10^5} \text{ F} $$

$$ C \approx 1.11 \times 10^{-6} \text{ F} = 1.11 \text{ } \mu\text{F} $$

Calculating Current at Resonance (\(I_0\))

At the resonant frequency (\(\omega_0\)), the impedance of the series RLC circuit is purely resistive and is equal to R. The source voltage is V. The current (\(I_0\)) at resonance is given by Ohm's law:

$$ I_0 = \frac{\text{Source Voltage (V)}}{\text{Impedance at Resonance (R)}} $$

$$ I_0 = \frac{V}{R} $$

We are given the source voltage, \(V = 100 \text{ V}\), and we calculated the resistance, \(R = 30 \text{ } \Omega\).

$$ I_0 = \frac{100 \text{ V}}{30 \text{ } \Omega} = \frac{10}{3} \text{ A} $$

Calculating Voltage Across Capacitor (\(V_C\)) at Resonance

The voltage across the capacitor (\(V_C\)) is the product of the current (\(I_0\)) flowing through it and its capacitive reactance (\(X_C\)) at the resonant frequency (\(\omega_0\)).

$$ V_C = I_0 \times X_C = I_0 \times \frac{1}{\omega_0 C} $$

We know \(I_0 = \frac{V}{R}\) and from the resonance condition \(\omega_0 = \frac{1}{\sqrt{LC}}\), we can rearrange to get \(\frac{1}{\omega_0 C} = \omega_0 L\).

Substituting these into the \(V_C\) formula:

$$ V_C = \left(\frac{V}{R}\right) \times (\omega_0 L) $$

$$ V_C = \frac{V \omega_0 L}{R} $$

We have:

• Source Voltage, \(V = 100 \text{ V}\)
• Resonant Frequency, \(\omega_0 = 3000 \text{ rad/sec}\)
• Inductance, \(L = 0.1 \text{ H}\)
• Resistance, \(R = 30 \text{ } \Omega\)

Substitute these values into the formula for \(V_C\):

$$ V_C = \frac{100 \text{ V} \times 3000 \text{ rad/sec} \times 0.1 \text{ H}}{30 \text{ } \Omega} $$

$$ V_C = \frac{100 \times 300}{30} \text{ V} $$

$$ V_C = \frac{30000}{30} \text{ V} $$

$$ V_C = 1000 \text{ V} $$

So, the voltage across the capacitor at resonance is 1000 V.

Summary of Results

Based on the calculations:

• The value of resistance, R, is 30 \(\Omega\).
• The voltage across the capacitor, \(V_C\), at resonance is 1000 V.

Therefore, the value of R and the voltage across C are, respectively, 30 \(\Omega\) and 1000 V.

Revision Table: Series RLC Circuit Formulas

Additional Information on RLC Circuits

Series RLC circuits are fundamental in electrical engineering, used in applications like filters, oscillators, and tuning circuits. Resonance is a key phenomenon where the circuit stores maximum energy in the electric and magnetic fields. The voltage across the capacitor and inductor can be significantly higher than the source voltage at resonance, a phenomenon known as voltage magnification or quality factor (Q). The quality factor Q is given by \(Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 RC}\), and the voltage across C or L at resonance is \(V_C = V_L = V \times Q\). In this problem, \(Q = \frac{3000 \text{ rad/sec} \times 0.1 \text{ H}}{30 \text{ } \Omega} = \frac{300}{30} = 10\). The voltage across C is \(V \times Q = 100 \text{ V} \times 10 = 1000 \text{ V}\), confirming our result.