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The percentage increase in the impedance of an ac circuit, when its power factor changes form 0.866 to 0.5 is (Resistance constant)
- a)73.2%
- b)86.6%
- c)90.8%
- d)66.6%
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The percentage increase in the impedance of an ac circuit, when its po...
Power factor (cosϕ)=R/Z1
For first case, power factor=0.866
Or, R/ Z1=0.866
Or,Z1=1.154R
For second case, power factor=>0.5
Cosϕ2=>R/Z2=>0.5, Z2=2R.
% increase=>[(Z2-Z1)/Z1] x100
So, [{(2R)-(1.154R)}/(1.154R)]x100=73.2%
So the correct answer is option A.
For first case, power factor=0.866
Or, R/ Z1=0.866
Or,Z1=1.154R
For second case, power factor=>0.5
Cosϕ2=>R/Z2=>0.5, Z2=2R.
% increase=>[(Z2-Z1)/Z1] x100
So, [{(2R)-(1.154R)}/(1.154R)]x100=73.2%
So the correct answer is option A.
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Community Answer
The percentage increase in the impedance of an ac circuit, when its po...
Explanation:
The formula for impedance in an AC circuit is given by:
Z = R + jX
where R is the resistance, X is the reactance, and j is the imaginary unit.
The power factor is given by:
cos Φ = R/Z
where Φ is the phase angle between the voltage and current.
When the power factor changes from 0.866 to 0.5, the angle Φ increases, which means that the reactance X also increases. However, the resistance R remains constant.
The percentage increase in impedance can be calculated as follows:
Let Z1 be the impedance at a power factor of 0.866, and let Z2 be the impedance at a power factor of 0.5.
Z1 = R + jX1
Z2 = R + jX2
cos Φ1 = R/Z1
cos Φ2 = R/Z2
Since the resistance R is constant,
cos Φ1 = cos Φ2
Therefore,
R/Z1 = R/Z2
Z2/Z1 = Z2(R)/Z1(R)
Z2/Z1 = (R + jX2)/(R + jX1)
Taking the modulus of both sides,
|Z2/Z1| = |(R + jX2)/(R + jX1)|
|Z2/Z1| = |(R + jX2)/(R + jX1)| x |(R - jX2)/(R - jX2)|
|Z2/Z1| = |(R^2 + X2^2)/(R^2 + X1^2)|
|Z2/Z1| = (R^2 + X2^2)/(R^2 + X1^2)
The percentage increase in impedance is given by:
% increase = (|Z2/Z1| - 1) x 100
% increase = [(R^2 + X2^2)/(R^2 + X1^2) - 1] x 100
Substituting X1 = R tan Φ1 and X2 = R tan Φ2,
% increase = [(R^2 + R^2 tan^2 Φ2)/(R^2 + R^2 tan^2 Φ1) - 1] x 100
% increase = [(1 + tan^2 Φ2)/(1 + tan^2 Φ1) - 1] x 100
% increase = [(1 + 1/3)/(1 + 3) - 1] x 100
% increase = 73.2%
Therefore, the percentage increase in impedance is 73.2%.
The formula for impedance in an AC circuit is given by:
Z = R + jX
where R is the resistance, X is the reactance, and j is the imaginary unit.
The power factor is given by:
cos Φ = R/Z
where Φ is the phase angle between the voltage and current.
When the power factor changes from 0.866 to 0.5, the angle Φ increases, which means that the reactance X also increases. However, the resistance R remains constant.
The percentage increase in impedance can be calculated as follows:
Let Z1 be the impedance at a power factor of 0.866, and let Z2 be the impedance at a power factor of 0.5.
Z1 = R + jX1
Z2 = R + jX2
cos Φ1 = R/Z1
cos Φ2 = R/Z2
Since the resistance R is constant,
cos Φ1 = cos Φ2
Therefore,
R/Z1 = R/Z2
Z2/Z1 = Z2(R)/Z1(R)
Z2/Z1 = (R + jX2)/(R + jX1)
Taking the modulus of both sides,
|Z2/Z1| = |(R + jX2)/(R + jX1)|
|Z2/Z1| = |(R + jX2)/(R + jX1)| x |(R - jX2)/(R - jX2)|
|Z2/Z1| = |(R^2 + X2^2)/(R^2 + X1^2)|
|Z2/Z1| = (R^2 + X2^2)/(R^2 + X1^2)
The percentage increase in impedance is given by:
% increase = (|Z2/Z1| - 1) x 100
% increase = [(R^2 + X2^2)/(R^2 + X1^2) - 1] x 100
Substituting X1 = R tan Φ1 and X2 = R tan Φ2,
% increase = [(R^2 + R^2 tan^2 Φ2)/(R^2 + R^2 tan^2 Φ1) - 1] x 100
% increase = [(1 + tan^2 Φ2)/(1 + tan^2 Φ1) - 1] x 100
% increase = [(1 + 1/3)/(1 + 3) - 1] x 100
% increase = 73.2%
Therefore, the percentage increase in impedance is 73.2%.
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