JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2024
- Hard
- Numerical
A capacitor of reactance \(4 \sqrt{3} \Omega\) and a resistor of resistance \(4 \Omega\) are connected in series with an ac source of peak value \(8 \sqrt{2} \mathrm{~V}\). The power dissipation in the circuit is _______ \(W\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{X}^2 \mathrm{~L}}\)
\(\mathrm{Z}=\sqrt{4^2+(4 \sqrt{3})^2}=8 \Omega\)
\(\mathrm{V}_{\mathrm{mms}}=\frac{\mathrm{V}}{\sqrt{2}}=\frac{8 \sqrt{2}}{\sqrt{2}}=(8 \mathrm{~V})\)
\(\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{z}}=\frac{8}{8}=1 \mathrm{~A}\)
Power dissipated \(=\mathrm{I}_{\mathrm{rms}}^2 \times \mathrm{R}=1 \times 4=(4 \mathrm{~W})\)

\(\mathrm{Z}=\sqrt{4^2+(4 \sqrt{3})^2}=8 \Omega\)
\(\mathrm{V}_{\mathrm{mms}}=\frac{\mathrm{V}}{\sqrt{2}}=\frac{8 \sqrt{2}}{\sqrt{2}}=(8 \mathrm{~V})\)
\(\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{z}}=\frac{8}{8}=1 \mathrm{~A}\)
Power dissipated \(=\mathrm{I}_{\mathrm{rms}}^2 \times \mathrm{R}=1 \times 4=(4 \mathrm{~W})\)

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