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A resistance of 50W, an inductance of 20/p henry and a capacitor of 5/p mF are connected in series with an A.C. source of 230 volt and 50Hz. The impedance of circuit is
  • a)
    5W
  • b)
    50W
  • c)
    5KW
  • d)
    500W
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A resistance of 50W, an inductance of 20/p henry and a capacitor of 5/...
Given, R=50W, L=(20/π)H and C=(5/π)mF
Voltage supply, V=230V.
v=50Hz
Since, XL= ω=2πvL=2πx50x(20/π)=XL=2000 Ω
And, XC= 1/ωC=1/2πvL=1/[2πx50x(5/π)x10-6]=(1000/500) x103
So, impedance, Z=√[(XL-XC)2+R2]
Z=√[(2000-2000)2+502]
Z=50Ω
Hence, Option B is correct.
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Most Upvoted Answer
A resistance of 50W, an inductance of 20/p henry and a capacitor of 5/...
Calculation of Impedance

To calculate the impedance of the circuit, we use the formula:

Z = √(R² + (Xl - Xc)²)

Where R is the resistance, Xl is the inductive reactance and Xc is the capacitive reactance.

Applying the values given in the question, we get:

R = 50Ω
Xl = 2πfL = 20/πΩ
Xc = 1/(2πfC) = 1/(2π*50*5/10^9) = 63.66Ω

Substituting the values in the formula, we get:

Z = √(50² + (20/π - 63.66)²)
= √(2500 + 16.44²)
= √(2500 + 270.99)
= √2770.99
= 52.63Ω

Therefore, the impedance of the circuit is 52.63Ω, which is closest to option B (50Ω).
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A resistance of 50W, an inductance of 20/p henry and a capacitor of 5/p mF are connected in series with an A.C. source of 230 volt and 50Hz. The impedance of circuit isa)5Wb)50Wc)5KWd)500WCorrect answer is option 'B'. Can you explain this answer?
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