Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Simple A.C. circults
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Simple A.C. circults
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Subject: Physics
Class: Senior Secondary 3
Term: 1st Term
Week: 1
Theme: Fields At Rest And In Motion
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Students should be able to:- explain the peak and v.m.s values of current and the p.d. establish the phase relationship between current and p.d in an a.c. circuit Explain reactance and impedance.
Determine current in circuits containing:- Resistance and capacitance- Resistance, in ductance, and capacitance.- Determine power in an a.c. circuit.
Alternating current (A.C.) and voltage vary sinusoidally with time. Peak Value ($I_0$, $V_0$): This is the maximum value reached by the alternating current or voltage in one cycle. It represents the amplitude of the sinusoidal waveform. Mathematically, for an A.C. voltage: $V = V_0 \sin(\omega t)$ For an A.C. current: $I = I_0 \sin(\omega t)$ where $V_0$ and $I_0$ are the peak voltage and peak current, respectively. R.M.
S. Value ($I_{rms}$, $V_{rms}$): The Root Mean Square (r.m.s.) value of A.C. current or voltage is the effective value that produces the same heating effect in a resistor as a steady D.C. current or voltage of the same magnitude. It is the value usually quoted for A.C. supplies (e.g., 240V, 50Hz in Nigeria refers to $V_{rms}$). For a sinusoidal waveform, the r.m.s. value is related to the peak value by: $I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$ $V_{rms} = \frac{V_0}{\sqrt{2}} \approx 0.707 V_0$ Significance: R.M.S. values are crucial because they allow for direct comparison between A.C. and D.C. power delivery in terms of heating or power consumption.
Worked Example 1: The mains supply in Nigeria is typically 240 V, 50 Hz. Determine the peak voltage and peak current if a 100 W filament bulb (purely resistive) is connected to the supply.
Solution: Given: $V_{rms} = 240 \text{ V}$, $f = 50 \text{ Hz}$, Power $P = 100 \text{ W}$. Determine Peak Voltage ($V_0$): $V_{rms} = \frac{V_0}{\sqrt{2}}$ $V_0 = V_{rms} \times \sqrt{2} = 240 \text{ V} \times 1.414 = 339.36 \text{ V}$ The peak voltage is approximately 339.4
V. Determine Peak Current ($I_0$): First, find the r.m.s. current ($I_{rms}$) using the power formula for a purely resistive circuit: $P = V_{rms} I_{rms}$. $I_{rms} = \frac{P}{V_{rms}} = \frac{100 \text{ W}}{240 \text{ V}} = 0.4167 \text{ A}$ Now, use the relationship between $I_{rms}$ and $I_0$: $I_0 = I_{rms} \times \sqrt{2} = 0.4167 \text{ A} \times 1.414 = 0.589 \text{ A}$ The peak current is approximately 0.589 A. The phase relationship describes the timing difference between the voltage and current waveforms. This difference is expressed as a phase angle ($\phi$). a)
Purely Resistive Circuit (R): When an A.C. voltage is applied across a pure resistor, the current flowing through it is always in phase with the voltage across it. This means they both reach their peak, zero, and minimum values at the same instant. If $V = V_0 \sin(\omega t)$, then $I = I_0 \sin(\omega t)$. Phase angle $\phi = 0^\circ$.
Phasor Diagram: Voltage and current phasors are drawn along the same direction. b)
Purely Inductive Circuit (L): When an A.C. voltage is applied across a pure inductor (coil with negligible resistance), the changing magnetic flux induces a back e.m.f. that opposes the change in current. This causes the current to lag behind the voltage by a phase angle of $90^\circ$ (or $\pi/2$ radians). If $V = V_0 \sin(\omega t)$, then $I = I_0 \sin(\omega t - \pi/2)$. Phase angle $\phi = +90^\circ$ (voltage leads current).
Phasor Diagram: Voltage phasor leads the current phasor by $90^\circ$ counter-clockwise. c)
Purely Capacitive Circuit (C): When an A.C. voltage is applied across a pure capacitor, the capacitor charges and discharges, causing current to flow. The current is maximum when the voltage is changing most rapidly (at zero voltage) and zero when the voltage is maximum. This means the current leads the voltage by a phase angle of $90^\circ$ (or $\pi/2$ radians). If $V = V_0 \sin(\omega t)$, then $I = I_0 \sin(\omega t + \pi/2)$. Phase angle $\phi = -90^\circ$ (current leads voltage).
Phasor Diagram: Current phasor leads the voltage phasor by $90^\circ$ counter-clockwise.
Reactance: This is the opposition to A.C. current flow offered by inductors and capacitors. Unlike resistance, reactance depends on the frequency of the A.C. supply. It is measured in Ohms ($\Omega$). Inductive Reactance ($X_L$): The opposition offered by an inductor to the flow of A.C. current. It is directly proportional to the inductance (L) and the angular frequency ($\omega$). $X_L = \omega L = 2\pi f L$ where $f$ is the frequency in Hertz (Hz) and $L$ is the inductance in Henry (H). As frequency increases, $X_L$ increases. At D.C. ($f=0$), $X_L=0$, so an inductor acts as a short circuit. Capacitive Reactance ($X_C$): The opposition offered by a capacitor to the flow of A.C. current. It is inversely proportional to the capacitance (C) and the angular frequency ($\omega$). $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$ where $f$ is the frequency in Hertz (Hz) and $C$ is the capacitance in Farad (F). As frequency increases, $X_C$ decreases. At D.C. ($f=0$), $X_C = \infty$, so a capacitor acts as an open circuit.
Impedance (Z): This is the total effective opposition to the flow of A.C. current in a circuit containing resistors, inductors, and capacitors. It is the A.C. equivalent of D.C. resistance. Impedance is also measured in Ohms ($\Omega$). For a series R-L-C circuit, resistance (R) is in phase with voltage, $X_L$ causes voltage to lead current by $90^\circ$, and $X_C$ causes current to lead voltage by $90^\circ$. Because $X_L$ and $X_C$ have opposite phase effects, their effects tend to cancel each other out. The impedance Z is the vector sum of R, $X_L$, and $X_C$.
Impedance for an R-L-C Series Circuit: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ Impedance for an R-C Series Circuit: (where $L=0$, so $X_L=0$) $Z = \sqrt{R^2 + X_C^2}$ Impedance for an R-L Series Circuit: (where $C=0$, so $X_C=0$) $Z = \sqrt{R^2 + X_L^2}$ Phase Angle ($\phi$): The phase angle between the total voltage and total current in the circuit is given by: $\tan \phi = \frac{X_L - X_C}{R}$ If $\phi > 0$, the circuit is inductive (voltage leads current). If $\phi < 0$, the circuit is capacitive (current leads voltage). If $\phi = 0$, the circuit is purely resistive (voltage and current are in phase).
Worked Example 2: A 200 $\Omega$ resistor, a 0.5 H inductor, and a 10 $\mu F$ capacitor are connected in series to a 240 V, 50 Hz A.C. supply.
Calculate: a) Inductive reactance b) Capacitive reactance c) Impedance of the circuit d) The current flowing in the circuit e) The phase angle between the voltage and current Solution: Given: $R = 200 \Omega$, $L = 0.5 \text{ H}$, $C = 10 \mu F = 10 \times 10^{-6} \text{ F}$, $V_{rms} = 240 \text{ V}$, $f = 50 \text{ Hz}$. a) Inductive Reactance ($X_L$): $X_L = 2\pi f L = 2 \times 3.142 \times 50 \text{ Hz} \times 0.5 \text{ H} = 157.1 \Omega$ b) Capacitive Reactance ($X_C$): $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.142 \times 50 \text{ Hz} \times 10 \times 10^{-6} \text{ F}} = \frac{1}{0.003142} = 318.3 \Omega$ c)
Impedance (Z): $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $Z = \sqrt{200^2 + (157.1 - 318.3)^2}$ $Z = \sqrt{40000 + (-161.2)^2}$ $Z = \sqrt{40000 + 25985.44} = \sqrt{65985.44} = 256.9 \Omega$ d) Current ($I_{rms}$): Using Ohm's Law for A.C. circuits: $I_{rms} = \frac{V_{rms}}{Z}$ $I_{rms} = \frac{240 \text{ V}}{256.9 \Omega} = 0.934 \text{ A}$ e) Phase Angle ($\phi$): $\tan \phi = \frac{X_L - X_C}{R} = \frac{157.1 - 318.3}{200} = \frac{-161.2}{200} = -0.806$ $\phi = \arctan(-0.806) = -38.87^\circ$ The negative sign indicates that the current leads the voltage (capacitive circuit). In an A.C. circuit, the instantaneous power ($P$) delivered to a component or circuit varies with time.
However, what is generally of interest is the average power delivered over a complete cycle. Average Power ($P_{avg}$): In an A.C. circuit, only the resistive components dissipate actual average power. Inductors and capacitors store and release energy, but ideally do not dissipate average power. $P_{avg} = V_{rms} I_{rms} \cos \phi$ where $\cos \phi$ is the power factor. Power Factor ($\cos \phi$): This is the cosine of the phase angle ($\phi$) between the voltage and current. It indicates how effectively the power is being utilized in the circuit. For a purely resistive circuit, $\phi = 0^\circ$, $\cos \phi = 1$. $P_{avg} = V_{rms} I_{rms}$ (maximum power transfer). For a purely inductive or capacitive circuit, $\phi = \pm 90^\circ$, $\cos \phi = 0$. $P_{avg} = 0$ (no average power dissipated). For an R-L-C circuit, $\cos \phi = \frac{R}{Z}$.
Therefore, $P_{avg} = V_{rms} I_{rms} \frac{R}{Z} = I_{rms}^2 Z \frac{R}{Z} = I_{rms}^2 R$. This confirms that power is only dissipated in the resistance.
Worked Example 3: Using the circuit details from Worked Example 2 (200 $\Omega$ resistor, 0.5 H inductor, 10 $\mu F$ capacitor in series with 240 V, 50 Hz supply). Calculate the average power dissipated in the circuit.
Solution: From Worked Example 2, we have: $V_{rms} = 240 \text{ V}$ $I_{rms} = 0.934 \text{ A}$ $\phi = -38.87^\circ$ $R = 200 \Omega$ Method 1: Using Power Factor $\cos \phi = \cos(-38.87^\circ) = 0.7787$ $P_{avg} = V_{rms} I_{rms} \cos \phi = 240 \text{ V} \times 0.934 \text{ A} \times 0.7787 = 174.4 \text{ W}$ Method 2: Using $I_{rms}^2 R$ $P_{avg} = I_{rms}^2 R = (0.934 \text{ A})^2 \times 200 \Omega = 0.872356 \times 200 = 174.47 \text{ W}$ Both methods yield approximately the same result. The average power dissipated is about 174.5 W.
Power Transmission and Distribution in Nigeria: The national grid (managed by Transmission Company of Nigeria, TCN, and distributed by DISCOs) primarily uses A.C. This is because A.C. voltage can be easily stepped up (for efficient long-distance transmission with minimal power loss) and stepped down (for safe household and industrial use) using transformers. Understanding impedance and power factor is critical for minimizing energy losses during transmission, which is a major challenge for consistent electricity supply in Nigeria.
Electronic Appliances and Gadgets: Most household appliances in Nigeria (refrigerators, fans, air conditioners, televisions) operate on A.C. Different components within these appliances (motors, power supplies) utilize A.C. principles, including capacitance and inductance, for their operation and filtering purposes. For example, a fan motor is an inductive load, and understanding its phase relationship helps in its design and power consumption.
Radio and Communication Systems: RLC circuits are fundamental to tuning radios and other communication devices. The phenomenon of resonance (where $X_L = X_C$) allows a radio receiver to select a specific frequency (station) from many incoming radio waves, a common technology used in local radio stations across Nigeria.
Generators and Inverters: Many homes and businesses in Nigeria rely on A.C. generators and inverters during power outages. These devices produce A.C. electricity, and their operation, efficiency, and safety features are directly linked to the principles of A.C. circuits, including voltage regulation, frequency control, and current output.