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Electric Current Physics Current Electricity Physics Class 12th

A uniform heating wire of resistance 36 $Ω$ is connected across a potential difference of 240V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is________.

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alok kumar singh | Contributor-Level 10

2 months ago

In first condition R₁ = 36 $Ω$

In second condition R₂ = 18 $Ω$

$P_{1} = \frac{V^{2}}{R_{1}} = \frac{{(240)}^{2}}{36}$

$P_{2} = \frac{V^{2}}{R_{2}} + \frac{V^{2}}{R_{2}} = \frac{{(240)}^{2}}{18} + \frac{{(240)}^{2}}{18}$

$P_{2} = \frac{{(240)}^{2}}{9}$

So $\frac{P_{1}}{P_{2}} = \frac{{(240)}^{2} / 36}{{(240)}^{2} / 9} = \frac{1}{4}$

x = 4

In first condition R1 = 36 <math> <mrow> <mi>Ω</mi> </mrow> </math>  In second condition R2 = 18  <math> <mrow> <mi>Ω</mi> </mrow> </math> <math> <mrow> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mn>4</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> <mn>6</mn> </mrow> </mfrac> </mrow> </math>                <math> <mrow> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mn>4</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mn>4</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> </mrow> </math>                <math> <mrow> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mn>4</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> </mrow> </math>               So   <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mn>4</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>/</mo> <mn>3</mn> <mn>6</mn> </mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mn>4</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>/</mo> <mn>9</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math> x = 4

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A uniform heating wire of resistance 36 $Ω$ is connected across a potential difference of 240V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is________.

alok kumar singh

In first condition R₁ = 36 $Ω$

In second condition R₂ = 18 $Ω$

$P_{1} = \frac{V^{2}}{R_{1}} = \frac{{(240)}^{2}}{36}$

$P_{2} = \frac{V^{2}}{R_{2}} + \frac{V^{2}}{R_{2}} = \frac{{(240)}^{2}}{18} + \frac{{(240)}^{2}}{18}$

$P_{2} = \frac{{(240)}^{2}}{9}$

So $\frac{P_{1}}{P_{2}} = \frac{{(240)}^{2} / 36}{{(240)}^{2} / 9} = \frac{1}{4}$

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