Investigating Resonance in a Series RLC Circuit

10- RLC Circuits Abstract: In and RLC circuit, the frequency of the AC voltage affects the current passing through the circuit. We will study this effect to find the resonance frequency, the frequency in which the current is maximum. The resonance frequency is affected by both the capacitor C and the inductor (coil) L and not by the resistor. Introduction: The RLC circuit is composed of a resistor, an inductor, and a capacitor, all connected on series with an alternating current source, usually sinusoidal. The resistor has a resistance R, this resistance is not affected by the frequency of the source. The capacitor and the inductor on the other hand have what we call a capacitive reactance (Xc) and an inductive reactance (XL) respectively. Those reactance are affected by the frequency of the source. The capacitive reactance depends on the capacitance and the frequency. It decreases as either the capacitance or frequency increase. It has the formula: 𝑋𝐶 = 1 2𝜋𝑓𝐶 The Inductive reactance increases as either the frequency or the inductance of the coil (L). It has the formula: 𝑋𝐿 = 2𝜋𝑓𝐿 The inductance L depends on the physical properties of coil, such as the number of turns and the area encircled by the coil. Similarly, the capacitance C depends on the physical properties of the capacitor. A common parallel plate capacitor’s capacitance depends on the area of the plate, the separation of the plates, along the permittivity of the dielectric filling the separation. In a normal resistor, both the current and voltage are in phase. But that’s not the case for capacitors and inductors. In a capacitor, current leads the voltage with 90°. Meaning that there is a phase difference of 90° between the voltage of the capacitor and the current passing. In inductors, the voltage leads the current with a phase difference of 90°. We use phasor diagrams to demonstrate this. This means that the voltage across L and C have a 180° phase difference. We can treat them as vectors and find the voltage of the source. We get We can now calculate the source voltage Vin as: 𝑉𝑖𝑛 2 = (𝑉𝐿 − 𝑉𝐶 )2 + 𝑉𝑅 2 Or 𝑉𝑖𝑛 = √(𝑉𝐿 − 𝑉𝐶 )2 + 𝑉𝑅 2 This explains why we get higher voltages for both the capacitor and the inductor than the source. We can also find theta, the angle between VR, which is in the same direction as the current, using: 𝜃 = tan−1 𝑉𝐿 − 𝑉𝐶 𝑉𝑅 This theta is the phase difference between Vin and I. Now, to find the current, we need to find the impedance. Using: 𝑉𝐿 = 𝐼𝑋𝐿 , 𝑉𝐶 = 𝐼𝑋𝐶 , 𝑉𝑅 = 𝐼𝑅 : 𝑉𝑖𝑛 = √(𝐼𝑋𝐿 − 𝐼𝑋𝐶 )2 + (𝐼𝑅)2 𝑉𝑖𝑛 = 𝐼√(𝑋𝐿 − 𝑋𝐶 )2 + (𝑅)2 We define the impedance (Z) as 2 2 𝑍 = √(𝑋𝐿 − 𝑋𝐶 ) + (𝑅) . We now get Ohm’s Law: 𝑉𝑖𝑛 = 𝐼𝑍 Or 𝐼= 𝑉𝑖𝑛 𝑍 We will get the maximum current for the voltage when the impedance value falls to its lowest value. This happens at a certain frequency such that XL and XC are equal. At that frequency, called the resonance frequency, the impedance Z=R. 𝑋𝐶 = 𝑋𝐿 1 = 2𝜋𝑓𝐿 2𝜋𝑓𝐶 1 = (2𝜋𝑓)2 𝐿𝐶 𝑓= 1 2𝜋 ξ𝐿𝐶 Under this frequency, the current will be maximum. Band Width: the difference between the upper frequency and the lower frequency that result in current equal to 0.707 of Imax. Quality Factor: The ratio of fr to the band width. It shows the energy loss due to the quantity of energy contained in the design. Procedure: 1- Connect the AC source, the resistance box, the inductor coil, the capacitor, and an Ammeter and all in series 2- Set the AC signal to the highest amplitude, and the resistance box to 0 ohms. 3- Change the frequency until you get the highest reading on the ammeter. Take note of the frequency (this will be f r) and the current (Imax). 4- Increase the frequency until you get a value of I that is 10 mA less and note the frequency. 5- Repeat step 4 five times. The go back to fr and lower the frequency until you get current that is 10 mA lower. Repeat 5 times. 6- You can measure the value of Vin ,VR ,VL ,VC to verify the previously mentioned relationship. 7- Change the value of the resistance to study the effect on current. 8- Graph the f – I graph and find the band width. Data: f(kHz) I(mA) f (lower) (Hz) I (mA) f (higher) (Hz) 27.00 30 27.30 40 27.50 50 27700 80 27700 27.60 60 27650 70 27750 27.65 70 27.70 80 27600 60 27800 27.75 70 27500 50 27900 27.80 60 27.90 50 27300 40 28100 28.10 40 27000 30 28400 28.40 30 Sources of error: 1- Accuracy in measurement tools. 2- Human error. Not reading the measurement accurately. 3- Approximation. 𝐼𝑚𝑎𝑥 = 80 𝑚𝐴 𝑓𝑟 = 27.7 𝑘𝐻𝑧 = 1 2𝜋ξ𝐿𝐶 ξ𝐿𝐶 = 5.75 × 10−6 𝐿𝐶 = 3.31 × 10−11 𝐼𝑚𝑎𝑥 ξ2 = 56.57 𝑚𝐴 𝑓𝑙𝑜𝑤𝑒𝑟 = 27.570 𝑘𝐻𝑧 𝑓ℎ𝑖𝑔ℎ𝑒𝑟 = 27.830 𝑘𝐻𝑧 ∆𝑓 = 𝑏𝑎𝑛𝑑 𝑤𝑖𝑑𝑡ℎ = 27.57 − 27.83 = 0.26 𝑘𝐻𝑧 𝑄= 𝑓𝑟 = 106.5 ∆𝑓 Discussion and Conclusion: We were able to find the resonance frequency. We can use our findings to find the value of R given L, or the other way around and the value of the source’s voltage. The graph of current vs frequency is sharp, the band width is small. 1- The voltage leads current in the coil, it lags in the capacitor, and is in the same phase in the resistance. 2- This phase change results in a vector addition of voltage, this is why the voltage across the components of the circuit do not add up like scalers. This allows for the voltage across the L and C to be larger than the input voltage. 3- The impedance of a circuit changes with the change of the frequency of the source. 1 4- The smallest impedance occurs at fr when XR and XL are equal. 𝑓𝑟 = 2𝜋 ξ𝑅𝐿 5- The RLC circuit will pass current when the frequency of the source is around the resonance frequency best. If the frequency is much lower the capacitive reactance will increase resulting in small to no current. When the frequency is increased above fr the inductive reactance increases and passes small to no current as well. 6- The band width shows us the range of frequencies around f r that the circuit will pass with current being above 0.707 I max. 7- We can calculate the band width and the Q factor mathematically using the formula: 𝐿 ∆𝜔 = 2𝜋∆𝑓 = 𝑅 and 1 𝐿 𝑄 = 𝑅 √𝐶 8- As the resistance decreases, Imax decreases, but fr stays unaffected. ∆𝑉 9- If we had the source voltage, we could find R using 𝑅 = 𝐼𝑠𝑜𝑢𝑟𝑐𝑒 , then 𝑚𝑎𝑥 find the capacitance and the inductance.