Question
Download Solution PDFA 220V, 50 Hz ac source is connected in series to a 30 Ω resistor, an inductor, and a capacitor, each having 200 Ω inductive reactance and 160 Ω capacitive reactance, respectively. The voltage drop across the resistor is ______.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
\(R=30\Omega,X_L=200\Omega,X_C=160\Omega,V=220V\)
Concept:
Impedance is the opposition to the alternating current presented by the combined effect of resistance and reactance in a circuit.
Impedance is given by the formula,
- \(Z=\sqrt{R^2+(X_L-X_C)^2}\)
Where \(R\) is Resistance,
\(X_L\) is inductive reactance,
\(X_C\) is capacitive reactance
Explanation:
- \(Z=\sqrt{R^2+(X_L-X_C)^2}\)
- \(Z=\sqrt{30^2+(200-160)^2}=\sqrt{30^2+40^2}=\sqrt{2500}=50\Omega\)
- \(Z=\sqrt{30^2+40^2}=\sqrt{2500}=50\Omega\)
Current, I\(=\frac{V}{Z}=\frac{220}{50}=\frac{22}{5}A\)
Potential drop across the resistor is \(V_d=IR=\frac{22}{5}\times30=132V\)
The correct answer is Option-1-132V.
Last updated on May 26, 2025
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