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In a pure inductive circuit, if the supply frequency is reduced to 1/2, the current will :
- a)be four times as high
- b)be doubled
- c)be reduced by half
- d)be reduced to one fourth
Correct answer is option 'B'. Can you explain this answer?
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In a pure inductive circuit, if the supply frequency is reduced to 1/2...
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In a pure inductive circuit, if the supply frequency is reduced to 1/2...
Explanation:
In a pure inductive circuit, the current lags behind the voltage by 90 degrees. The relationship between the voltage and current in an inductive circuit is given by the equation:
V = IωL
Where:
V = Voltage
I = Current
ω = Angular frequency (2πf)
L = Inductance
When the supply frequency is reduced to 1/2, the angular frequency (ω) is also reduced by half. As a result, the voltage (V) remains constant, but the reactance (Xl) increases.
Impedance in an Inductive Circuit:
The impedance (Z) in an inductive circuit is given by the equation:
Z = √(R^2 + Xl^2)
Where:
R = Resistance
Xl = Inductive reactance
Since the resistance (R) remains constant, and Xl increases when the frequency is reduced, the impedance of the circuit increases.
Current in an Inductive Circuit:
The current (I) in an inductive circuit is given by the equation:
I = V/Z
As the impedance (Z) increases, the current (I) decreases. This is because the voltage (V) remains constant, but the impedance (Z) increases, resulting in a smaller current.
Effect of Frequency Reduction:
When the supply frequency is reduced to 1/2, the impedance of the circuit increases, and as a result, the current decreases. The current is directly proportional to the voltage and inversely proportional to the impedance. Therefore, when the frequency is reduced, the current in a pure inductive circuit is reduced.
Answer:
Therefore, the correct answer is option 'B' - the current will be doubled.
In a pure inductive circuit, the current lags behind the voltage by 90 degrees. The relationship between the voltage and current in an inductive circuit is given by the equation:
V = IωL
Where:
V = Voltage
I = Current
ω = Angular frequency (2πf)
L = Inductance
When the supply frequency is reduced to 1/2, the angular frequency (ω) is also reduced by half. As a result, the voltage (V) remains constant, but the reactance (Xl) increases.
Impedance in an Inductive Circuit:
The impedance (Z) in an inductive circuit is given by the equation:
Z = √(R^2 + Xl^2)
Where:
R = Resistance
Xl = Inductive reactance
Since the resistance (R) remains constant, and Xl increases when the frequency is reduced, the impedance of the circuit increases.
Current in an Inductive Circuit:
The current (I) in an inductive circuit is given by the equation:
I = V/Z
As the impedance (Z) increases, the current (I) decreases. This is because the voltage (V) remains constant, but the impedance (Z) increases, resulting in a smaller current.
Effect of Frequency Reduction:
When the supply frequency is reduced to 1/2, the impedance of the circuit increases, and as a result, the current decreases. The current is directly proportional to the voltage and inversely proportional to the impedance. Therefore, when the frequency is reduced, the current in a pure inductive circuit is reduced.
Answer:
Therefore, the correct answer is option 'B' - the current will be doubled.
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In a pure inductive circuit, if the supply frequency is reduced to 1/2, the current will :a)be four times as highb)be doubledc)be reduced by halfd)be reduced to one fourthCorrect answer is option 'B'. Can you explain this answer?
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