PHY 102 Week 10 Alternating Current PDF

PHY 102 - GENERAL PHYSICS II: WEEK 10 (2-Credit Units) Department of Physics and Materials Science. Kwara State University, Malete, Nigeria. Lecturer: A. A. SHOLAGBERU...

PHY 102 - GENERAL PHYSICS II: WEEK 10 (2-Credit Units) Department of Physics and Materials Science. Kwara State University, Malete, Nigeria. Lecturer: A. A. SHOLAGBERU July 16, 2024 Course Outline: AC voltages and currents applied to Inductor, capacitors, and resistance. Course Objective: At the end of the lesson, the students will learn: 1. The meaning of Alternating Current and Phasors 2. Current and Potential relations. 3. Series L-R circuit. 4. Series C-R circuit. 5. Series L-R-C circuit. Contents 1 Introduction 3 2 Production of Alternating current 3 3 Application of alternating Current 3 4 Average value of AC and rms value of AC wave 3 5 Phasors Diagram 4 6 AC circuit containing Resistance only 5 7 Capacitor in an AC circuit 6 8 Inductor in an AC circuit 6 9 Impedance 7 10 Series L-R Circuit 7 11 Series C-R Circuit 8 12 Series L-C-R Circuit 8 13 Worked Examples 8 1 CONTENTS CONTENTS 14 Assignment 9 2  4 AVERAGE VALUE OF AC AND RMS VALUE OF AC WAVE 1 Introduction An alternating current can be defined as a current that changes its magnitude and polarity at regular intervals of time. A time varying current or voltage may be periodic and non-periodic. In case of periodic current or voltage, the current or voltage is said to be alternating if its amplitude is constant and alternate half cycle is positive and half negative. If the current or voltage varies periodically as sine or cosine function of time, the current or voltage is said to be sinusoidal. 2 Production of Alternating current Alternating current can be produced or generated by using devices that are known as alternators. An alternator is a device that converts mechanical energy into AC electrical energy. The distinctive difference between an alternator and a generator is that generator can generate either AC or DC while alternator only generates AC. The basic principle of the AC generator is a direct consequence of Faraday’s law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency ω = 2πf a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage and current are given as: V = V0 sin(ωt) i = i0 sin(ωt) 3 Application of alternating Current AC is the form of current mostly used in different appliances. Some of the examples of alternating current usage include audio signal, radio signal, etc. An alternating current has a wide advantage over DC because AC enables power to be transmitted over large distances with minimal loss of energy. AC is also capable of powering electric motors that further convert electrical energy into mechanical energy. Due to this, AC also finds its use in many large appliances like refrigerators, dishwashers and many other appliances. 4 Average value of AC and rms value of AC wave The average value of an ac is usually defined as the average of the instantaneous values of alternating current over a complete cycle. If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the 3  5 PHASORS DIAGRAM quantity to be measured varies with time. The average value of current for one cycle is ∫T [i0 sin(ωt)]dt < i >One−cycle = (iav )T = 0 ∫ T =0 0 dt Similarly, the average value of the voltage (or emf) for one cycle is zero. < V >One−cycle = 0 Since, these averages for the whole cycle are zero, the DC instrument will indicate zero deflection. In AC, the average value of current is defined as its average taken over half the cycle. Hence, ∫T 2 [i0 sin(ωt)]dt 2 < i >Half cycle = (iav ) T = 0 ∫T = i0 ≈ 0.637i0 2 2 dt π 0 for the average emf: 2 < V >Half cycle = V0 ≈ 0.637V0 π Root mean square Value RMS value is defined as the square root of means of squares of instantaneous values. It can also be described as the amount of AC power that generates the same heating effect as an equivalent DC power. i0 irms = √ ≈ 0.707i0 2 V0 Vrms = √ ≈ 0.707V0 2 The square root of the mean square value is called the virtual value and is the value given by AC instruments. Thus, when we speak of our house hold power supply as 220 V AC, this means that the rms voltage is 220 V and its voltage amplitude is √ V0 = 2Vrms = 311.13V (1) Form factor is the ratio of the rms value and the average value √ rmsv alue V0 2 π F ormf actor = = = √ = 1.11 (2) averagev alue 2V0 /π 2 2 5 Phasors Diagram The phasor diagram is used to determine the phase relationships between two or more sine waves propagating with the same frequency. Here, we use the terms lead, lag and also in-phase, out-of-phase to indicate the relation between one waveform with the other. To simplify our analysis of circuits containing two or more elements, we use graphical constructions called phasor diagrams. In these constructions, alternating (sinusoidal) quantities, such as current and voltage are rotating vectors called phasors. In these diagrams, the instantaneous value of a quantity that varies sinusoidally with time is represented by the projection onto a vertical axis (if it is a sine function) or onto a horizontal axis (if it is a cosine function) of a vector with a length equal to the amplitude (i0 ) of the quantity. The vector rotates counterclockwise with constant angular velocity ω. 4  6 AC CIRCUIT CONTAINING RESISTANCE ONLY 6 AC circuit containing Resistance only The pure resistive AC circuit contains only pure resistance of R ohms. There will be no effect of inductance and capacitance in this circuit. The alternate current and voltage move along both directions as backwards and forwards. Therefore, current and voltage follow a shape of sine. Figure 1: Phasor diagram for a resistor in an AC circuit Consider a resistor with resistance R through which there is a sinusoidal current given by VR = iR = (i0 R) sin ωt where V0 = i0 R is the voltage amplitude 5  8 INDUCTOR IN AN AC CIRCUIT 7 Capacitor in an AC circuit If a capacitor of capacitance C is connected across the alternating source, the instantaneous charge on the capacitor is q = CVC = CV0 sin ωt the instantaneous current i passing through the given circuit is: dq i= = CV0 ω cos ωt dt i0 V0 = ωC 1 This relation shows that the quantity ωC is the effective AC resistance or the capacitive reactance of the capacitor and is represented as XC. It has unit as ohm. Thus, 1 XC = ωC Figure 2: Phasor diagram for a capacitor in an AC circuit It is clear that the current leads the voltage by 90ř or the potential drop across the capacitor lags the current passing it by 90ř. In another word, the voltage phasor is behind the current phasor by a quarter cycle or 90ř. 8 Inductor in an AC circuit Consider a pure inductor of self-inductance L and zero resistance connected to an alternating source. Again we assume that an instantaneous current i = i0 sin ωt flows through the inductor. Although, there is no resistance, there is a potential difference VL between the inductor terminals a and b because the current varies with time giving rise to a self-induced emf. 6  10 SERIES L-R CIRCUIT Figure 3: Phasor diagram for a Inductor in an AC circuit di VL = L = Li0 ω cos ωt dt V0 i= sin ωt ωL the inductive reactance XL is XL = ωL the maximum current, V0 i0 = XL From equation above, we see that the voltage across the inductor leads the current passing through it by 90ř. 9 Impedance Impedance is a combination of resistance and reactance. It is essentially anything and everything that obstructs the flow of electrons within an electrical circuit. Hence, it affects the generation of current through the electrical circuit. It is present in all the possible components of the circuit and across all possible electrical circuits. Impedance is mathematically symbolized by the letter Z and has its unit as ohm. It is a superset of both resistance and reactance combined. In phasor terms, impedance Z is represented as a combination of resistance R and reactance X. 10 Series L-R Circuit As we have established, potential difference across a resistance in AC is in phase with current and it leads the current by 90ř in phase. Here, the impedance Z = R + jXL. its modulus is given as: √ |Z| = R2 + (ωL)2 7  13 WORKED EXAMPLES the potential difference leads the current by an angle, ( ) −1 ωL ϕ = tan R 11 Series C-R Circuit Potential difference across a capacitor in AC lags in phase by 90ř with the current in the circuit. Suppose in phasor diagram current is taken along positive x-direction. Then, VR is also along positive x-direction but VC is along negative y-direction. So, we can write the impedance as 1 Z = R + j ωC. its modulus is given as: √ ( )2 1 |Z| = R +2 ωC the potential difference lags the current by an angle, ( ) −1 1 ϕ = tan ωRC 12 Series L-C-R Circuit Potential difference across an inductor leads the current by 90ř in phase while that across a capacitor, it lags in phase by 90ř. Suppose in a phasor diagram current is taken along positive x-direction. Then, VR is along positive x-direction, VL along positive y-direction and VC along negative y-direction. Here the impedance is: ( ) 1 Z = R + j(XL + XC ) = R + j ωL + ωC √ ( )2 1 |Z| = R2 + ωL − ωC and the potential difference leads the current by an angle, ( 1 ) −1 ωL − ωC ϕ = tan R 13 Worked Examples Example 1: If the current in an AC circuit is represented by the equation, i = 5 sin(300t − π4 ) Here, t is in second and i in ampere. Calculate (a) peak and rms value of current (b) frequency of AC (c) average current solution (a) i = 5 sin(300t − π4 ) The peak value, i0 = 5 A the rms value is irms = √i02 = √52 = 3.535 A (b) Angular frequency, ω = 300 rad/s ∴ f = 2π ω = 300 2π ≈ 47.75 8  14 ASSIGNMENT (2) (2) (c) iav = π i0 = π (5) = 3.18 A Example 2: A 100 Ω resistance is connected in series with a 4 H inductor. The voltage across the resistor is VR = 2.0 Vsin(103 rad/s)t : (a) Find the expression of circuit current (b) Find the inductive reactance (c) Derive an expression for the voltage across the inductor. Solution 3 )t (a) i = VRR = 2.0V sin((10 100 = 2.0−2 V sin((103 rad/s)t (b)XL = ΩL = 103 rad/s ×4 H = 4.0 × 103 Ω (c) The amplitude of the voltage across inductor V0 = i0 XL = (2.0−2 ) A 4.0 × 103 Ω = 80 V π In an AC, voltage across the inductor leads the current by 90ř or 2 rad. Hence, VL = V0 sin(ωt + π/2) = 80 sin(103 t + π/2) Example 3 Find the voltage across the various elements, i.e. resistance, capacitance and inductance which√ are in series and having values 1000 Ω, 1 µF and 2.0 H, respectively. Given emf is V = 100 2 sin 100t volt. solution The rms value √ of the voltage across the source, √ 2 = 100 V Vrms = 100 2 ω = 1000 rad/s irms = V|Z| rms = √ 2 Vrms = √ 2 Vrms 1 2 R +(XL −XC ) 2 R +(ωL− ωC ) √ 100 = 1 10002 +(1000×2− −6 ) 2 1000×1×10 = 0.0707 A 14 Assignment Theoretical Questions Question 1 A choke coil is needed to operate an arc lamp at 160 V (rms) and 50 Hz. The lamp has an effective resistance of 5 Ω when running at 10 A (rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160 V (DC ), what additional resistance is required? Compare the power loses in both cases. Question 2 A 300 resistor, a 0.250 H inductor, and a 8.00 ţF capacitor are in series with an AC source with voltage amplitude 120 V and angular frequency 400 rad/ s. (a) What is the current amplitude? (b) What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? (c) What are the voltage amplitudes across the resistor, inductor, and capacitor. Question 3 A 5.00 H inductor with negligible resistance is connected across an AC source. Voltage amplitude is kept constant at 60.0 V but whose frequency can be varied. Find the current amplitude when the angular frequency is 9  14 ASSIGNMENT (a) 100 rad/s (b) 1000 rad/s (c) 10000 rad/s Multiple choice Questions 1. The term cos ϕ in an AC circuit is called (a) form factor (b) phase factor (c) power factor (d) quality factor 2. A DC ammeter cannot measure alternating current because (a) AC changes its direction (b) DC instruments will measure the average value (c) AC can damage the DC instrument (d) AC produces more heat 3. The rms value of an alternating current (a) √ is equal to 0.707 times peak value (b) is equal to 0.636 times peak value (c) is equal to 2 times the peak value (d) None of the above 4. A generator produces a time varying voltage given by V = 240 sin 120t , where t is in second. The rms voltage and frequency are (a) 170 V and 19 Hz (b) 240 V and 60 Hz (c) 170 V and 60 Hz (d) 120 V and 19 Hz 5. The frequency of an alternating current is 50 Hz. The minimum time taken by it in reaching from zero to peak value is (a) 5 ms (b) 10 ms (c) 20 ms (d) 50 ms 10