Electrical Engineering: Analysis of Electric Iron and Power Calculation, Study notes of Electrical Engineering

Download Electrical Engineering: Analysis of Electric Iron and Power Calculation and more Study notes Electrical Engineering in PDF only on Docsity! 1. State Ohm’s law Ans- Ohm's law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, given a constant temperature. Mathematically, it can be expressed as 𝐼=𝑉𝑅I=RV, where 𝐼I is the current in amperes (A), 𝑉V is the voltage in volts (V), and 𝑅R is the resistance in ohms (ΩΩ). 2. Limitation of Ohm's Law. Ans-Ohm's Law has limitations, particularly in non-linear devices like diodes and transistors where the relationship between current and voltage is not linear. Additionally, it assumes constant temperature and linear materials, which may not always hold true in practical circuits. 3. State Kirchoff’s Current law with circuit diagram. Ans - Kirchhoff's Current Law (KCL) states that the total current entering a junction in an electrical circuit is equal to the total current leaving the junction. This is based on the principle of conservation of charge 4. State Kirchoff’s volatge law with circuit diagram Ans- Kirchhoff's Voltage Law (KVL) states that the sum of all electrical potential differences (voltages) around any closed loop in a circuit is zero. This law is based on the principle of conservation of energy. 5. 5 -An Electric iron is rated 1000W, 240V. Find the current drawn & resistance of the heating element. Ans- To find the current drawn and the resistance of the heating element of an electric iron rated at 1000W and 240V, we can use the following electrical formulas: Current Drawn (I) The power formula relating power (P), voltage (V), and current (I) is: 𝑃=𝑉×𝐼P=V×I Rearranging this formula to solve for current (I): 𝐼=𝑃𝑉I=VP Substituting the given values: 𝐼=1000 W240 VI=240V1000W 𝐼≈4.17 AI≈4.17A Resistance of the Heating Element (R) Using Ohm's Law, which relates voltage (V), current (I), and resistance (R): 𝑉=𝐼×𝑅V=I×R Rearranging this formula to solve for resistance (R): 𝑅=𝑉𝐼R=IV Substituting the values we found for voltage and current: 𝑅=240 V4.17 AR=4.17A240V 𝑅≈57.55 ΩR≈57.55Ω Summary  Current drawn: 𝐼≈4.17 AI≈4.17A  Resistance of the heating element: 𝑅≈57.55 ΩR≈57.55Ω 6 – An Electric iron is rated 1000W, 240V operating for 2hr. Find the energy disscipated by the heating element. Ans - To find the energy dissipated by the heating element of an electric iron rated at 1000W operating for 2 hours, we can use the formula for electrical energy: Energy=Power×TimeEnergy=Power×Time Given:  Power 𝑃=1000 WP=1000W  Time 𝑡=2 hourst=2hours First, we need to convert the time from hours to seconds to use standard SI units (optional, but we'll keep it simple in hours): Energy=1000 W×2 hoursEnergy=1000W×2hours Since 1 watt-hour (Wh) is equal to 3600 joules (J), we can convert the energy to joules: Energy=1000×2 Wh=2000 WhEnergy=1000×2Wh=2000Wh Converting to joules: 2000 Wh×3600 JWh=7,200,000 J2000Wh×3600WhJ=7,200,000J Summary The energy dissipated by the heating element is 2000 Wh2000Wh or 7,200,000 J7,200,000J. 7 - Write down the expression of equivalent resistance for ‘n’ – number of resistors in series connection. Ans- For 𝑛n resistors connected in series, the equivalent resistance 𝑅eqReq is the sum of all individual resistances. Expression: 𝑅eq=𝑅1+𝑅2+𝑅3+⋯+𝑅𝑛Req=R1+R2+R3+⋯+Rn Where 𝑅1,𝑅2,𝑅3,…,𝑅𝑛R1,R2,R3,…,Rn are the resistances of the individual resistors. 8 - Write down the expression of equivalent resistance for ‘n’ – number of resistors in parallel connection. Ans - The expression for the equivalent resistance (𝑅𝑒𝑞Req) of 'n' resistors in parallel connection is: 𝑅𝑒𝑞=11𝑅1+1𝑅2+1𝑅3+…+1𝑅𝑛Req=R11+R21+R31+…+Rn11 Where 𝑅1,𝑅2,𝑅3,…,𝑅𝑛R1,R2,R3,…,Rn are the resistances of the individual resistors. 9- Differentiate between loop and mesh. Ans- In the context of electrical circuits:  Line voltage refers to the voltage between any two lines or conductors in the system. It is typically higher in magnitude compared to phase voltage and is denoted as 𝑉𝐿VL.  Phase voltage refers to the voltage across each individual phase of the system. In a balanced three-phase system, the phase voltages are identical in magnitude and are denoted by 𝑉phVph. 20- What is meant by Real power? Ans- Real power, often denoted as 𝑃P, is the actual power consumed or transferred by an electrical circuit, device, or system, measured in watts (W). It represents the energy transferred per unit of time, which is effectively converted into useful work, such as mechanical movement, heat, or light. 21- What is meant by apparent power Ans - Apparent power is the combination of real power and reactive power in an AC circuit, representing the total power flow, measured in volt-amperes (VA). 22 - What is reactive power? Ans - Reactive power is the portion of apparent power in an AC circuit that is due to the phase difference between voltage and current, causing energy to oscillate between the source and load without being consumed, measured in volt-amperes reactive (VAR). 23- What are the limitations of Thevenin’s theorem? Ans- The limitations of Thevenin's theorem include: 1. It's applicable only to linear networks. 2. It's limited to networks with only voltage and current sources (no dependent sources). 3. The theorem assumes the network to be in a steady-state condition. 4. It's not valid for networks containing non-linear elements like diodes and transistors. 24- What are the limitations of Norton's theorem? Ans - The limitations of Norton's theorem include: 1. It's applicable only to linear networks. 2. It's limited to networks with only voltage and current sources (no dependent sources). 3. The theorem assumes the network to be in a steady-state condition. 4. It's not valid for networks containing non-linear elements like diodes and transistors. 25- What are the limitations of Superposition theorem? Ans - The limitations of the Superposition theorem include: 1. It applies only to linear circuits. 2. It's restricted to circuits with only independent sources (no dependent sources). 3. It assumes that each source must be analyzed separately, which can be time-consuming for complex circuits. 4. It doesn't consider interactions between sources, so it may not accurately predict behavior in circuits with strong interactions between components. 26- Define active elements and passive elements. Ans - Active elements are components that can generate energy or amplify signals, such as voltage and current sources, transistors, and amplifiers. Passive elements are components that do not generate energy but can only dissipate, store, or control it, such as resistors, capacitors, and inductors. 27- Define current. Ans - Current is the flow of electric charge per unit of time through a conductor, measured in amperes (A). 28- Two resistances of 4 ohm and 6 ohms are connected in parallel across 10v battery. Determine the current through 6-ohm resistance. Ans - To find the current through the 6-ohm resistance, we first need to find the equivalent resistance of the parallel combination of the 4-ohm and 6-ohm resistors. 1𝑅eq=1𝑅1+1𝑅2Req1=R11+R21 1𝑅eq=14+16Req1=41+61 1𝑅eq=312+212Req1=123+122 1𝑅eq=512Req1=125 𝑅eq=125Req=512 𝑅eq=2.4 ΩReq=2.4Ω Now, we can use Ohm's Law to find the current through the 6-ohm resistance: 𝐼=𝑉𝑅I=RV 𝐼=106I=610 𝐼=1.67 AI=1.67A So, the current through the 6-ohm resistance is 1.67 A1.67A. 29 - Write down the expression for effective resistance when three resistances are connected in series and parallel. Ans -Sure, here are the expressions: For resistances connected in series: 𝑅series=𝑅1+𝑅2+𝑅3Rseries=R1+R2+R3 For resistances connected in parallel: 1𝑅parallel=1𝑅1+1𝑅2+1𝑅3Rparallel1=R11+R21+R31 30- State advantages of 3Ø System Over1Ø System? Ans - The advantages of a three-phase (3Ø) system over a single-phase (1Ø) system include higher power transmission efficiency, smoother power delivery, lower conductor size requirement, and the ability to support balanced loads without a neutral wire. 31- Draw Delta Connected 3-Ø supply system. Ans - 32- Draw star Connected 3-Ø supply system. Ans- 33- Three inductive coils each with resistance of 15Ω and an inductance of 0.03H are connected in star to a 3 phase 400V, 50Hz supply. Calculate the phase voltage. Ans - To calculate the phase voltage (𝑉phVph) in a star-connected system, use the following relationship between line voltage (𝑉lineVline) and phase voltage: 46- Write definition and Formula of Power factor. Ans- Power factor is the ratio of real power (in watts) to apparent power (in volt-amperes) in an AC circuit, representing the efficiency of power usage and indicating the phase relationship between voltage and current. Formula: Power factor=Real power (W)Apparent power (VA)Power factor=Apparent power (VA)Real power (W) 47-Explain impedance triangle. Ans - The impedance triangle is a graphical representation used in AC circuits to visualize the relationship between resistance (R), inductive reactance (X_L), and capacitive reactance (X_C), forming a right triangle where the impedance (Z) is the hypotenuse. It helps understand the phase relationship between voltage and current in an AC circuit.