AP EAMCET · PHYSICS · Alternating Current
An inductor of reactance \(1 \Omega\) and a resistor of resistance \(3 \Omega\) are connected in series to the terminals of \(10 \mathrm{~V}\) (rms) AC source. The power dissipated in the circuit is
- A \(33.3 \mathrm{~W}\)
- B \(30 \mathrm{~W}\)
- C \(31.6 \mathrm{~W}\)
- D \(20 \mathrm{~W}\)
Answer & Solution
Correct Answer
(B) \(30 \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
Average power dissipated. \(P_{\mathrm{avg}}=\frac{v_{\mathrm{rms}}^2 R}{z^2}\) Here, \(Z=\) impedence \(=\sqrt{X_L^2+R^2}\) where, \(X_L=\) inductive reactance \(=L \omega\) So impedence, \(z=\sqrt{X_L^2+R^2}\) \(=\sqrt{1^2+3^2}\) (Here given, \(X_L=1 \Omega\) and…
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