Laboratory No. 3 Series RLC Circuit, Lab Reports of Electrical Circuit Analysis

Download Laboratory No. 3 Series RLC Circuit and more Lab Reports Electrical Circuit Analysis in PDF only on Docsity! Laboratory Activity No. 3 AC Circuits Series RLC Circuit Introduction Resistance and impedance both represent opposition to the flow of the alternating current. Both are measured in terms of the same unit, the ohm. To determine the magnitude of the total impedance, get the sum of the impedance of each of the elements in series. As long as all the necessary calculations are carried out by vector algebra, use the two relationships studied earlier under DC circuits. The total impedance may not always increase with the addition of another element in series. Capacitive reactance could cancel out inductive reactance and vice versa. An extreme case would have the capacitive reactance completely cancelling out the inductive reactance. This results in resonance high voltages and current could result. Objectives 1. Determine the resistance, inductance and capacitance by a voltage and current measurements. 2. Compare measured and calculated voltages and current for a series RLC circuit. 3. Determine the impedance of a given circuit to compare the measure value to calculated value of a given circuit. Materials Multisim Circuit Diagrams Figure 1 Procedure: 1. Build the circuit given in figure 1 on Multisim. 2. Measure the total current and the current across the lamp, across 2.5372 H inductor and across the 5uF capacitor. To measure the total current, refer to figure 2a. Record the current reading at table 1. Repeat the steps for the current lamp, 2.5372 H inductor and 5uF capacitor. Refer to figure 2b, 2c and 2d. Figure 2a. Total Current 5. Using Ohm’s Law, compute the voltage and current for each component. Record it at Table 1 and 2. Use the formula 𝑅 = 𝑒𝑅 𝑖𝑅 ; 𝑋𝐿 = 𝑒𝐿 𝑖𝐿 ; 𝑋𝐶 = 𝑒𝑐 𝑖𝑐 𝑍 = 𝑉𝑡 𝐼𝑡 6. Compute the percent difference between the measure and the computed value of the impedance. Table 1: Simulation and Computation Result of Current for Figure 2 and 3 Voltage (V) R L C Total Workbench 0.67908 A 0.67908 A 0.67908 A 0.67908 A Computation 0.68242 A 0.68242 A 0.68242 A 0.68242 A % Difference 0.4920 % 0.4920 % 0.4920 % 0.4920 % Table 2: Simulation and Computation Result of Voltages for Figure 2 and 3 Current (A) R L C Total Workbench 328.76 V 651.63 V 359.10 V 439.29 V Computation 330.2918 V 652.7371V 362.0356 V 440 V % Difference 0.4934 % 0.1699 % 0.8175 % 0.1616 % Questions: 1. Do the workbench and computational values of voltages and currents agree? The workbench or the manual computation differ from each other by little. It produced only less than 1% of difference. 2. Give possible reasons for any discrepancies. Since the current is AC, the value accumulated in Multism changes from time to time and makes it difficult to capture the same value as the manually-computed one. Manual Computation Using Ohm’s Law Given: 𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 (𝑳) = 2.5327 𝐻 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒 (𝑪) = 5𝜇𝐹 = 5𝑥10−6 𝐹 Solution: 𝐶𝑢𝑟𝑟𝑒𝑛𝑡: 𝑰 = 𝑃 𝑉 = 100 𝑊 220 𝑉 𝑰 = 𝟓 𝟏𝟏 𝑨 𝐼𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒: 𝑿𝐿 = 2𝜋𝑓𝐿 = 2𝜋(60𝐻𝑧)(2.5327𝐻) 𝑿𝑳 = 𝟗𝟓𝟔. 𝟓𝟎𝟏𝟖𝟔𝟓𝟕 𝜴 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐶𝑖𝑟𝑐𝑢𝑖𝑡: 𝒁 = √𝑅2 + (𝑋𝐿 − 𝑋𝑐)2 𝒁 = √(484 Ω)2 + (956.5018657 Ω − 530.516477 Ω)2 𝒁 = 𝟔𝟒𝟒. 𝟕𝟔𝟑𝟏𝟕𝟒𝟕 𝜴 𝑇𝑜𝑡𝑎𝑙 𝐶𝑖𝑐𝑢𝑖𝑡𝑠′𝐶𝑢𝑟𝑟𝑒𝑛𝑡: 𝑰𝒕 = 𝑉 𝑍 = 220 𝑉 644.7631747 𝛺 𝑰𝒕 = 𝟎. 𝟑𝟒𝟏𝟐𝟏𝟎𝟓𝟓𝟑𝟗 𝑨 ≈ 𝟑𝟒𝟏. 𝟐𝟏𝟎𝟓𝟓𝟑𝟗 𝒎𝑨 Since it is in Series, the value of current that flows around within the circuit is the same. 𝑐𝑜𝑠 𝜃 = 𝑅 𝑍 → 𝜃 = 𝑐𝑜𝑠−1 484 𝛺 644.7631747 𝛺 ∴ 𝜽 = 𝟒𝟏. 𝟑𝟓𝟐𝟕𝟑𝟗𝟑𝟔° 𝒍𝒂𝒈𝒈𝒊𝒏𝒈 𝑃 = 100 𝑊 𝑉 = 220 𝑉 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒: 𝑿𝐶 = 1 2𝜋𝑓𝐶 = 1 2𝜋(60 𝐻𝑧)(5𝑥10−6 𝐹) 𝑿𝑪 = 𝟓𝟑𝟎. 𝟓𝟏𝟔𝟒𝟕𝟕 𝜴 𝑃𝑒𝑎𝑘 𝑡𝑜 𝑃𝑒𝑎𝑘 𝑉𝑎𝑙𝑢𝑒: 𝑰𝒑𝒑 = (341.2105539 𝑚𝐴)(2) 𝑰𝒑𝒑 = 𝟔𝟖𝟐. 𝟒𝟐𝟏𝟏𝟎𝟕𝟖 𝒎𝑨 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑢𝑙𝑏: 𝑹 = 𝑉2 𝑃 = (220 𝑉)2 100 𝑊 𝑹 = 𝟒𝟖𝟒 𝛀 Manual Computation Using the Formula 𝑽𝒐𝒍𝒕𝒆𝒔 𝒂𝒄𝒓𝒐𝒔𝒔 𝒕𝒉𝒆 𝑺𝒆𝒓𝒊𝒆𝒔 𝑹𝑳𝑪 𝑪𝒊𝒓𝒄𝒖𝒊𝒕: 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑏𝑢𝑙𝑏 (𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒) 𝒆𝑹 = (𝐼𝑡)(𝑅) = (0.3412105539 𝐴)(484Ω) 𝒆𝑹 = 𝟏𝟔𝟓. 𝟏𝟒𝟓𝟗𝟎𝟖𝟏 𝑽 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟 (𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒) 𝒆𝑳 = (𝐼𝑡)(𝑋𝐿) = (0.3412105539 𝐴)(956.5018657 𝛺) 𝒆𝑳 = 𝟑𝟐𝟔. 𝟑𝟔𝟖𝟓𝟑𝟏𝟒 𝑽 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟 (𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒) 𝒆𝑪 = (𝐼𝑡)(𝑋𝐶) = (0.3412105539 𝐴)(530.516477 𝛺) 𝒆𝑪 = 𝟏𝟖𝟏. 𝟎𝟏𝟕𝟖𝟐𝟏 𝑽 𝑇𝑜𝑡𝑎𝑙 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑽𝒕 = (𝐼𝑡)(𝑍) = (0.3412105539 𝐴)(644.7631747Ω) 𝑽𝒕 = 𝟐𝟐𝟎 𝑽 𝑍 = 𝑉𝑡 𝐼𝑡 𝑅 = 𝑒𝑅 𝑖𝑅 ; 𝑋𝐿 = 𝑒𝐿 𝑖𝐿 ; 𝑋𝐶 = 𝑒𝑐 𝑖𝑐 𝑃𝑒𝑎𝑘 𝑡𝑜 𝑃𝑒𝑎𝑘 𝑉𝑎𝑙𝑢𝑒: 𝒆𝑹𝒑𝒑 = (165.1459081 𝑉)(2) 𝒆𝑹𝒑𝒑 = 𝟑𝟑𝟎. 𝟐𝟗𝟏𝟖𝟏𝟔𝟐 𝑽 𝑃𝑒𝑎𝑘 𝑡𝑜 𝑃𝑒𝑎𝑘 𝑉𝑎𝑙𝑢𝑒: 𝒆𝑳𝒑𝒑 = (326.3685314 𝑉)(2) 𝒆𝑳𝒑𝒑 = 𝟔𝟓𝟐. 𝟕𝟑𝟕𝟎𝟔𝟐𝟗 𝑽 𝑃𝑒𝑎𝑘 𝑡𝑜 𝑃𝑒𝑎𝑘 𝑉𝑎𝑙𝑢𝑒: 𝒆𝑪𝒑𝒑 = (181.017821 𝑉)(2) 𝒆𝑪𝒑𝒑 = 𝟏𝟖𝟏. 𝟎𝟏𝟕𝟖𝟐𝟏 𝑽 𝑃𝑒𝑎𝑘 𝑡𝑜 𝑃𝑒𝑎𝑘 𝑉𝑎𝑙𝑢𝑒: 𝑽𝑻𝒑𝒑 = (181.017821 𝑉)(2) 𝑽𝑻𝒑𝒑 = 𝟏𝟖𝟏. 𝟎𝟏𝟕𝟖𝟐𝟏 𝑽 vy 164.13V Vpp 439.29V Vaus 155.56 Vay -190.80pV L 2.5372H c1 Ts v 136.51V Vpp 326.67V Vams 116.20V L 2.5372H Vv 220V 60Hz or ci Sur v 64.007V Vpp 651.63V ~ Vems 230.39V ow) ei aVav -18.602pV fy 60.000Hz 220v \O } L cp5372H | — cl Sur x 220V, 68.942V 2Vep2'359.10V Vams 126.96V Vay 143.39pv =-/fy 60.000Hz C41 SpF