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Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).?
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Two coils A and B are connected in series across a 240v, 50Hz supply. ...
Calculation of Impedance:
To calculate the impedance of the circuit, we need to consider the resistance and reactance of the coils A and B.
Resistance of Coil A:
Given that the resistance of coil A is 5 ohms.
Reactance of Coil B:
The reactance of a coil is given by the formula X = 2πfL, where X is the reactance, f is the frequency, and L is the inductance.
In this case, the frequency is 50Hz and the inductance of coil B is 0.02H. Substituting the values into the formula, we get X = 2π(50)(0.02) = 6.28 ohms.
Impedance:
The impedance (Z) of a series circuit is given by the formula Z = √(R^2 + X^2), where R is the resistance and X is the reactance.
Substituting the values, we have Z = √(5^2 + 6.28^2) = √(25 + 39.46) = √64.46 = 8.03 ohms.
Power Factor:
The power factor (PF) of a circuit is the cosine of the angle between the voltage and current waveforms. It is given by the formula PF = cos(θ), where θ is the phase angle.
Given that the power factor is 0.75, we can find the phase angle (θ) using the formula cos^(-1)(PF) = θ. Substituting the value, we have cos^(-1)(0.75) = θ, which gives θ = 41.41 degrees.
Power Consumed:
The power consumed (P) in a circuit is given by the formula P = VIcos(θ), where V is the voltage and I is the current.
Given that the voltage is 240V and the power consumed is 3kW (3000W), we can rearrange the formula to find the current (I): I = P/(Vcos(θ)) = 3000/(240cos(41.41)) ≈ 17.63A.
Summary:
- Resistance of coil A = 5 ohms.
- Reactance of coil B = 6.28 ohms.
- Impedance of the circuit = 8.03 ohms.
- Power factor = 0.75.
- Power consumed = 3kW.
- Current flowing through the circuit = 17.63A.
To calculate the impedance of the circuit, we need to consider the resistance and reactance of the coils A and B.
Resistance of Coil A:
Given that the resistance of coil A is 5 ohms.
Reactance of Coil B:
The reactance of a coil is given by the formula X = 2πfL, where X is the reactance, f is the frequency, and L is the inductance.
In this case, the frequency is 50Hz and the inductance of coil B is 0.02H. Substituting the values into the formula, we get X = 2π(50)(0.02) = 6.28 ohms.
Impedance:
The impedance (Z) of a series circuit is given by the formula Z = √(R^2 + X^2), where R is the resistance and X is the reactance.
Substituting the values, we have Z = √(5^2 + 6.28^2) = √(25 + 39.46) = √64.46 = 8.03 ohms.
Power Factor:
The power factor (PF) of a circuit is the cosine of the angle between the voltage and current waveforms. It is given by the formula PF = cos(θ), where θ is the phase angle.
Given that the power factor is 0.75, we can find the phase angle (θ) using the formula cos^(-1)(PF) = θ. Substituting the value, we have cos^(-1)(0.75) = θ, which gives θ = 41.41 degrees.
Power Consumed:
The power consumed (P) in a circuit is given by the formula P = VIcos(θ), where V is the voltage and I is the current.
Given that the voltage is 240V and the power consumed is 3kW (3000W), we can rearrange the formula to find the current (I): I = P/(Vcos(θ)) = 3000/(240cos(41.41)) ≈ 17.63A.
Summary:
- Resistance of coil A = 5 ohms.
- Reactance of coil B = 6.28 ohms.
- Impedance of the circuit = 8.03 ohms.
- Power factor = 0.75.
- Power consumed = 3kW.
- Current flowing through the circuit = 17.63A.
Community Answer
Two coils A and B are connected in series across a 240v, 50Hz supply. ...
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Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).?
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Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).?.
Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two coils A and B are connected in series across a 240v, 50Hz supply. The resistance of A is 5ohm and inductance of B is 0.02H. The power factor is 0.75. The impedance of the circuit is ( if power consumed is 3kW).?.
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