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- State Ohmโs 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 ๐ผ=๐๐
I=RV, where ๐ผI is the current in amperes (A), ๐V is the voltage in volts (V), and ๐
R is the resistance in ohms (ฮฉฮฉ).
- 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.
- State Kirchoffโs 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
- State Kirchoffโs 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 -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: ๐=๐ร๐ผP=VรI Rearranging this formula to solve for current (I): ๐ผ=๐๐I=VP Substituting the given values: ๐ผ=1000 W240 VI=240V1000W ๐ผโ4.17 AIโ4.17A Resistance of the Heating Element (R) Using Ohm's Law, which relates voltage (V), current (I), and resistance (R): ๐=๐ผร๐
V=IรR Rearranging this formula to solve for resistance (R): ๐
=๐๐ผR=IV Substituting the values we found for voltage and current: ๐
=240 V4.17 AR=4.17A240V ๐
โ57.55 ฮฉRโ57.55ฮฉ Summary ๏ท Current drawn: ๐ผโ4.17 AIโ4.17A ๏ท Resistance of the heating element: ๐
โ57.55 ฮฉRโ57.55ฮฉ 6 โ 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=PowerรTimeEnergy=PowerรTime Given: ๏ท Power ๐=1000 WP=1000W ๏ท Time ๐ก=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 Wร2 hoursEnergy=1000Wร2hours Since 1 watt-hour (Wh) is equal to 3600 joules (J), we can convert the energy to joules: Energy=1000ร2 Wh=2000 WhEnergy=1000ร2Wh=2000Wh Converting to joules: 2000 Whร3600 JWh=7,200,000 J2000Whร3600WhJ=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 โnโ โ number of resistors in series connection. Ans- For ๐n resistors connected in series, the equivalent resistance ๐
eqReq is the sum of all individual resistances. Expression: ๐
eq=๐
1+๐
2+๐
3+โฏ+๐
๐Req=R 1 +R 2 +R 3 +โฏ+Rn Where ๐
1,๐
2,๐
3,โฆ,๐
๐R 1 ,R 2 ,R 3 ,โฆ,Rn are the resistances of the individual resistors. 8 - Write down the expression of equivalent resistance for โnโ โ number of resistors in parallel connection. Ans - The expression for the equivalent resistance (๐
๐๐Req) of 'n' resistors in parallel connection is: ๐
๐๐=11๐
1+1๐
2+1๐
3+โฆ+1๐
๐Req=R 11 +R 21 +R 31 +โฆ+Rn 11 Where ๐
1,๐
2,๐
3,โฆ,๐
๐R 1 ,R 2 ,R 3 ,โฆ,Rn are the resistances of the individual resistors. 9- Differentiate between loop and mesh. Ans- In the context of electrical circuits:
For impedances (for impedance networks): ๐1=๐๐+๐๐+๐๐Z 1 =Za+Zb+Zc ๐2=๐๐+๐๐+๐๐Z 2 =Za+Zb+Zc ๐3=๐๐+๐๐+๐๐Z 3 =Za+Zb+Zc Where ๐๐Za, ๐๐Zb, and ๐๐Zc are the impedances in the star network. 17 - Write down the formula for a delta connected network is converted into a star network? Ans- To convert a delta (ฮ) connected network into a star (Y) network, you can use the following formulas: For resistances: ๐
๐=๐
1ร๐
2 ๐
1+๐
2+๐
3 Ra=R 1 +R 2 +R 3 R 1 รR 2 ๐
๐=๐
2ร๐
3 ๐
1+๐
2+๐
3 Rb=R 1 +R 2 +R 3 R 2 รR 3 ๐
๐=๐
3ร๐
1 ๐
1+๐
2+๐
3 Rc=R 1 +R 2 +R 3 R 3 รR 1 Where ๐
1 R 1 , ๐
2 R 2 , and ๐
3 R 3 are the resistances in the delta network. For impedances (for impedance networks): ๐๐=๐1ร๐ 2 ๐1+๐2+๐ 3 Za=Z 1 +Z 2 +Z 3 Z 1 รZ 2 ๐๐=๐2ร๐ 3 ๐1+๐2+๐ 3 Zb=Z 1 +Z 2 +Z 3 Z 2 รZ 3 ๐๐=๐3ร๐ 1 ๐1+๐2+๐ 3 Zc=Z 1 +Z 2 +Z 3 Z 3 รZ 1 Where ๐ 1 Z 1 , ๐ 2 Z 2 , and ๐ 3 Z 3 are the impedances in the delta network.
18- Define line currents and phase currents?
Ans - In three-phase electrical systems: ๏ท Line currents refer to the currents flowing through the lines or conductors connecting the power source to the load. These currents are typically higher in magnitude compared to phase currents and are represented by ๐ผ๐ฟIL. ๏ท Phase currents refer to the currents flowing through each individual phase of the system. In a balanced three-phase system, the phase currents are identical in magnitude and are denoted by ๐ผphIph. 19- Define line voltage and phase voltage? Ans - In three-phase electrical systems:
๏ท 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 ๐๐ฟVL. ๏ท 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 ๐phVph. 20- What is meant by Real power? Ans- Real power, often denoted as ๐P, 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 Theveninโs theorem? Ans- The limitations of Thevenin's theorem include:
- It's applicable only to linear networks.
- It's limited to networks with only voltage and current sources (no dependent sources).
- The theorem assumes the network to be in a steady-state condition.
- 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:
- It's applicable only to linear networks.
- It's limited to networks with only voltage and current sources (no dependent sources).
- The theorem assumes the network to be in a steady-state condition.
- 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:
- It applies only to linear circuits.
- It's restricted to circuits with only independent sources (no dependent sources).
- It assumes that each source must be analyzed separately, which can be time-consuming for complex circuits.
1 ๐
parallel=1๐
1+1๐
2+1๐
3 Rparallel 1 =R 11 +R 21 +R 31 30- State advantages of 3ร System Over1ร System? Ans - The advantages of a three-phase (3ร) system over a single-phase (1ร) 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-ร supply system. Ans - 32- Draw star Connected 3-ร supply system. Ans- 33- Three inductive coils each with resistance of 15ฮฉ 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 (๐phVph) in a star-connected system, use the following relationship between line voltage (๐lineVline) and phase voltage:
๐ph=๐line3Vph= 3 Vline Given: ๏ท Line voltage (๐lineVline) = 400V Now, calculate the phase voltage: ๐ph=400๐ 3 Vph= 3400 V ๐ph=400๐1.732Vph=1.732400V ๐phโ231๐Vphโ231V So, the phase voltage is approximately 231V 34- Three inductive coils each with resistance of 15ฮฉ and an inductance of 0.03H are connected in delata to a 3 phase 400V, 50Hz supply. Calculate the phase voltage. Ans - In a delta-connected system, the line voltage (๐lineVline) is equal to the phase voltage (๐phVph). Given: ๏ท Line voltage (๐lineVline) = 400V Therefore, the phase voltage is: ๐ph=๐line=400๐Vph=Vline=400V So, the phase voltage is 400V. 35- What are the three types of power used in a a.c circuit? Ans - The three types of power used in an AC circuit are:
- Real Power (P): The actual power consumed or used in the circuit, measured in watts (W).
- Reactive Power (Q): The power that oscillates between the source and reactive components, measured in volt-amperes reactive (VAR).
- Apparent Power (S): The total power in the circuit, combining both real and reactive power, measured in volt-amperes (VA). 36- Define average value. Ans- The average value is a measure of the central tendency of a set of values, calculated by adding up all the values and dividing by the total number of values. It provides an indication of the typical value of a dataset over a specified period of time or space. 37- Define reactive power. Ans- Reactive power is the component of apparent power in an AC circuit that oscillates between the source and reactive components without being consumed, measured in volt-amperes reactive (VAR). 38- Write and explain the general expression for AC signal.
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.
