Physics Alternating Current Alternating Current Overview Class 12th Physics

The current (i) at time t = 0 and t = $\infty$ respectively for the given circuit is:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>5</mn> <mi>E</mi> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mo>,</mo> <mfrac> <mrow> <mn>1</mn> <mn>8</mn> <mi>E</mi> </mrow> <mrow> <mn>5</mn> <mn>5</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>8</mn> <mi>E</mi> </mrow> <mrow> <mn>5</mn> <mn>5</mn> </mrow> </mfrac> <mo>,</mo> <mfrac> <mrow> <mn>5</mn> <mi>E</mi> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>5</mn> <mi>E</mi> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mo>,</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mi>E</mi> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mi>E</mi> </mrow> <mrow> <mn>3</mn> <mn>3</mn> </mrow> </mfrac> <mo>,</mo> <mfrac> <mrow> <mn>5</mn> <mi>E</mi> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> </mrow> </math> </span></p>

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Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

At t = 0, no current flows through inductor.

$i = \frac{E}{\frac{6 * 9}{6 + 9}} = \frac{5 E}{18}$

At t = $\infty$ , current flows through circuit as if inductor is shorted.

$i = \frac{E}{\frac{5}{2} + \frac{4}{5}} = \frac{10 E}{33}$

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