Physics Alternating Current Class 12th Physics Power in AC Circuit

The net impedance of circuit (as shown in figure) will be:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>25</mn> <mtext> </mtext> <mi mathvariant="normal">W</mi> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>10</mn> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi mathvariant="normal">Ω</mi> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>15</mn> <mtext> </mtext> <mi mathvariant="normal">W</mi> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>5</mn> <mroot> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mi mathvariant="normal">Ω</mi> </math> </span></p>

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Answered by

Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 4

Detailed Solution:

$ω = 2 π f \Rightarrow ω = 100 π$

$Z = \sqrt[]{R^{2} + {(X_{L} - X_{C})}^{2}}$

$= \sqrt[]{10^{2} + {(ω L - \frac{1}{ω C})}^{2}}$

$= \sqrt[]{100 + {(100 π * \frac{50}{π} * 10^{- 3} - \frac{1}{100 π * \frac{10^{3}}{π} * 10^{- 6}})}^{2}}$

$= \sqrt[]{100 + (5 - 10)^{2}}$

$= \sqrt[]{100 + 25}$

$Z = 5 \sqrt[]{5} Ω$

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