As the switch S is closed in the circuit shown in figure, current passed through it is

 

Correct Option - 1


Detailed Solution:

Let V  be the potential of the junction as shown in figure. Applying junction law, we have

or $\frac{20-V}{2}+\frac{5-V}{4}\overline{\to }\frac{V-0}{2}$

or $40-2V+5-V=2V$ or $5V=45\Rightarrow V=9V$

$\therefore i_3=\frac{V}{2}=4.5A$

<p>Let V&lt;!- [if gte mso 9]>&lt;xml&gt; <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1819783820"> </o:OLEObject>&lt;/xml&gt;&lt;! [endif]-&gt;&nbsp;be the potential of the junction as shown in figure. Applying junction law, we have</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1758258198phpMoa63f.jpeg" alt="" width="212" height="115" loading="lazy"></picture></div></div><p>or <span contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mn>0</mn> <mo>−</mo> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>5</mn> <mo>−</mo> <mi>V</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mover accent="true"> <mo>→</mo> <mo>¯</mo> </mover> <mfrac> <mrow> <mi>V</mi> <mo>−</mo> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> </span></p><p>or <span contenteditable="false"> <math> <mrow> <mn>4</mn> <mn>0</mn> <mo>−</mo> <mn>2</mn> <mi>V</mi> <mo>+</mo> <mn>5</mn> <mo>−</mo> <mi>V</mi> <mo>=</mo> <mn>2</mn> <mi>V</mi> </mrow> </math> </span> or <span contenteditable="false"> <math> <mrow> <mn>5</mn> <mi>V</mi> <mo>=</mo> <mn>4</mn> <mn>5</mn> <mo>⇒</mo> <mi>V</mi> <mo>=</mo> <mn>9</mn> <mi>V</mi> </mrow> </math> </span></p><p><span contenteditable="false"> <math> <mrow> <mo>∴</mo> <mo>&nbsp;</mo> <msub> <mrow> <mi>i</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>4</mn> <mo>.</mo> <mn>5</mn> <mi>A</mi> </mrow> </math> </span></p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment