Fluent Essays

Answering questionshttps://www.youtube.com/watch?v=o5HWQZhane4&ab_channel=KoryGoldammer https://www.youtube.com/watch?v=MFmsrtGrjXs&ab_channel=KoryGoldammer https://www.youtube.com/watch?v=0YLTuhecnJw&ab_channel=KoryGoldammer© N. B. Dodge 01/12 Lab 5 Worksheet Note: Lab 5 is generally the most challenging exercise in ENGR 2105. Please watch the videos and read Lab 5 carefully at least twice and then take your time on the exercises below to make sure that you understand the theoretical material. 1. In what quadrant of the complex plane are these numbers located? −12+j7 __________________ −10−j50 __________________ 8−j2 __________________ 1+j100 __________________ 2. Use the complex conjugate to convert the expressions to a real term and an imaginary term. 26 (6 − 𝑗4)⁄ __________________ (8 − 𝑗8) (2 + 𝑗2)⁄ __________________ 3. Inductor and capacitor impedances are given as: 𝑍𝐿 = 𝑗𝜔𝐿 and 𝑍𝐶 = 1 𝑗𝜔𝐶⁄ . Assume you have a 10μF capacitor and a 10mH inductor. Calculate the impedance of these components at the following frequencies and list in the space provided: 1 MHz (1,000,000 Hz): 𝑗𝜔𝐿 = _______________Ω 1 𝑗𝜔𝐶⁄ = _________ Ω 50KHz (50,000 Hz): 𝑗𝜔𝐿 = ______________ Ω 1 𝑗𝜔𝐶⁄ = ___________ Ω 0Hz: 𝑗𝜔𝐿 = _______________ Ω 1 𝑗𝜔𝐶⁄ = _______________ Ω 4. Different items in the time domain transform in different ways to the ω domain: Element Time Domain ω Domain Transform Applied Sinusoidal AC Voltage 𝑽𝒑 𝐜𝐨𝐬(𝝎𝒕) (Volts) Vp (Volts) Series Current 𝑰𝒑 𝐜𝐨𝐬(𝝎𝒕 + 𝜽) (Amps) 𝑰𝒑 (Amps) Resistance R (Ohms) R (Ohms) Inductance L (Henry’s) 𝒋𝝎𝑳 (Ohms) Capacitance C (Farads) 𝟏 𝒋𝝎𝑪⁄ (Ohms) © N. B. Dodge 01/12 Given a circuit with a Time domain values: 𝑣(𝑡) = 10 cos(1000𝑡) 𝑅 = 100 Ω L = 10 mH C = 10 μF Calculate the values in the ω domain for: Applied Voltage __________Volts Resistance __________Ω Impedance from Inductor __________ Ω Impedance from Capacitor __________ Ω 5. After transforming voltage and circuit to the ω domain, find the current by dividing voltage by impedance. This usually results in a complex number. To convert back to the time-domain, which is the answer sought, do four things: • Rationalize using complex conjugate; the result is an X ± jY representation. • Convert complex number to polar-coordinates: • The radial distance is the peak current. The angle is the phase angle. 6. Based on the procedure in 5 above, convert the following ω domain currents back into the time domain (assume ω = 1000): I = 10+j10 _____________ I = −8+j4 _____________ 7. If 𝑣(𝑡) = 10 cos(1000𝑡), R = 100 Ω, C = 100 μF , determine the expression for i(t) .© N. B. Dodge 01/12 ENGR 2105 – Inductors and Capacitors in AC Circuits and Phase Relationships 1. Introduction and Goal: Capacitors and inductors in AC circuits are studied. Impedance and phase relationships of AC voltage and current are defined. Frequency-dependence of inductor and capacitor impedance is introduced. Phase relationships of AC voltage and current are defined. 2. Equipment List: • Multisim • Scientific Calculator 3. Experimental Theory: Capacitors and inductors change the voltage- current relationship in AC circuits. Since most single-frequency AC circuits have a sinusoidal voltage and current, exercises in Experiment 5 use sinusoidal AC voltages. Note that in an RLC AC, current frequency will be identical to the voltage, although the current waveform will be different. “Imaginary” Numbers, the Complex Plane, and Transforms: 3.1.1 Definition of j: As √−1 is not a real number, Electrical Engineers define 𝑗 = +√−1. Physicists and Mathematicians use 𝑖 = −√−1 for this same purpose, so 𝑗 = −𝑖, but that will not affect our theory. 3.1.2 Electrical Engineering problem solutions often include imaginary numbers. It is useful to consider real and imaginary numbers as existing in a two dimensional space, one axis of which is a real-number axis, and the other of which is the “imaginary” axis. 3.1.3 In the complex plane (Figure 1), the horizontal axis is the real axis, and the vertical axis is the y axis. Real numbers (–7 , 10) lie on the x- axis, imaginary numbers (–3j, j42) lie on the y-axis. Complex numbers lie off axis. For example, 2 + j6 would lie in the first quadrant, and (–43 – j17) would lie in the third quadrant. © N. B. Dodge 01/12 3.1.4 Transforms: Transforms allow moving a problem from a coordinate system or domain where it is difficult to solve to one where it is easier to solve (Figure 2). 3.1.5 Simpler equations in the transform domain make the problem easier to solve than in the original domain. We can transfer sinusoidal, single-frequency AC circuit problems to a domain where we can use algebra to solve them rather than calculus. Solving problems in algebra is almost always easier than calculus! 3.1.6 The catch: We need transforms (formulas) to the new domain. Then “inverse transforms” are required to return the solution to the time domain, where it is useful. 3.2 A New Domain: In the phasor or frequency (ω) domain, sinusoidal AC circuit problems are easier to solve. Note: 𝛚 = 𝟐𝛑𝐟, where f is the frequency in Hertz. Recall that Hertz has units of 1/second, or “per second”. Thus, ω is in radians/sec. 3.2.1 Transforms: Skipping the derivation (you will do the derivation if you take ENGR 2305), we simply list frequency domain transforms. Note that AC voltage is © N. B. Dodge 01/12 usually expressed as 𝑣(𝑡) = 𝑉𝑝cos(ωt), where Vp is the peak voltage. 3.3 Transform from the Time Domain to the FreqExperiment #5 Data Sheet 1. RL Circuit measurements: 2.1.Measured peak (Vp) voltage of the Voltage Source: __________ 2.2.Peak current (A): __________ 2.3.Time delta (μsec) between current and voltage peaks: __________ 2.4.Time-domain expression for i(t), based on measures above: __________ 2.5.Calculated expression for i(t), based on measured R and L: __________ 2.6.List your ideas for discrepancies, if any: _________________________ 2. RC Circuit measurements: 2.1 Measured peak voltage of the Voltage Source: __________ 2.2 Peak current (A): __________ 2.3 Time delta (μsec) between current and voltage peaks: __________ 2.4 Time-domain expression for i(t), based on measures above: __________ 2.5 Calculated expression for i(t), based on measured R and C: __________ 2.6 List your ideas for discrepancies, if any: _________________________ _____________________________________________________________Brief 5 AC RL and RC Circuits Electrical Circuits Lab I (ENGR 2105) Dr. Kory Goldammer Review of Complex Numbers and Transforms Transforms The Polar Coordinates / Rectangular Coordinates Transform The Complex Plane We can use complex numbers to solve for the phase shift in AC Circuits Instead of (x,y) coordinates, we define a point in the Complex plane by (real, imaginary) coordinates Real numbers are on the horizontal axis Imaginary numbers are on the vertical axis The Complex Plane (cont.) Imaginary numbers are multiplied by j By definition, (Mathematicians use i instead of j, but that would confuse us since i stands for current in this class) Imaginary Plane: Rectangular Coordinates We can identify any point in the 2D plane using (real, imaginary) coordinates Complex Plane Using Rectangular Coordinates Imaginary Plane: Transform to Polar Coordinates We can identify any point in the complex plane using (r,) coordinates. The arrow is called a Phasor. r is the length of the Phasor, and is the angle between the Positive Real Axis and the Phasor is the Phase Angle we want to calculate Complex Plane Using Rectangular Coordinates r Imaginary Plane: Transform to Polar Coordinates Complex Plane Using Rectangular Coordinates (we will discuss the meaning of r later) Use either the sin or cos term to find : But we need in radians: r =7.07 Complex Math For addition or subtraction, add or subtract the real and j terms separately. (3 + j4) + (2 – j2) = 5 + j2 To multiply or divide a j term by a real number, multiply or divide the numbers. The answer is still a j term. 5 * j6 = j30 -2 * j3 = -j6 j10 / 2 = j5 Complex Math (cont. 1) To divide a j term by a j term, divide the j coefficients to produce a real number; the j factors cancel. j10 / j2 = 5 -j6 / j3 = -2 To multiply complex numbers, follow the rules of algebra, noting that j2 = -1 Complex Math (cont. 2) To divide by a complex number: Can’t be done! The denominator must first be converted to a Real number! Complex Conjugation Converting the denominator to a real number without any j term is called rationalization. To rationalize the denominator, we need to multiply the numerator and denominator by the complex conjugate Complex Number Complex Conjugate 5 + j3 5 – j3 –5 + j3 –5 – j3 5 – j3 5 + j3 –5 – j3 –5 + j3 Complex Math (cont. 2) Multiply the original equation by the complex conjugate divided by itself (again, j2 = -1): Phase Shift Time Domain - ω Domain Transforms Transforming from the Time (Real World) Domain to the (Problem Solving Domain Note that in the ω Domain, Resistance, Inductance and Capacitance all of units of Ohms! Element Time Domain ω Domain Transform Applied Sinusoidal AC Voltage (Volts) (ω=2πf) Vp (Volts) Series Current (Amps) (ω=2πf) (Amps) Resistance R (Ohms) R (Ohms) Inductance L (Henry’s) (Ohms) (ω=2πf) Capacitance