Two resistors $400\mathrm{\Omega }$  and   $800\mathrm{\Omega }$ are connected in series across a  $6\vee$ battery. The potential difference measured by a voltmeter of $10\mathrm{k}\mathrm{\Omega }$ across $400\mathrm{\Omega }$  resistor is close to:

Correct Option - 3


Detailed Solution:

Parallel of  $10\mathrm{k}\mathrm{\Omega }$ and $400\mathrm{\Omega }=384.61$ $\mathrm{\Omega }$

$\begin{array}{rr}{\mathrm{V}}_{400\mathrm{\Omega }}& =\frac{384.61}{384.61+800}\times 6\\ & =1.95\mathrm{V}\end{array}$

<p>Parallel of&nbsp; <span contenteditable="false"> <math> <mn>10</mn> <mi mathvariant="normal">k</mi> <mi mathvariant="normal">Ω</mi> </math> </span>and<span contenteditable="false"> <math> <mn>400</mn> <mi mathvariant="normal">Ω</mi> <mo>=</mo> <mn>384.61</mn> </math> </span><span contenteditable="false"> <math> <mi mathvariant="normal">Ω</mi> </math> </span></p><p><span contenteditable="false"> <math> <mtable columnalign="right"> <mtr> <mtd> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">V</mi> </mrow> </mrow> <mrow> <mrow> <mn>400</mn> <mi mathvariant="normal">Ω</mi> </mrow> </mrow> </msub> </mrow> </mtd> <mtd> <mrow> <mi> </mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>384.61</mn> </mrow> </mrow> <mrow> <mrow> <mn>384.61</mn> <mo>+</mo> <mn>800</mn> </mrow> </mrow> </mfrac> <mo>×</mo> <mn>6</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <mi> </mi> <mo>=</mo> <mn>1.95</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </mrow> </mtd> </mtr> </mtable> </math> </span></p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1756292795phptCYhPi.jpeg" alt="" width="312" height="236" loading="lazy"></picture></div></div>

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