A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of $3 Ω a n d 2 Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:

Option 1 -

3 : 2

Option 2 -

3 : 1

Option 3 -

3 : 4

Option 4 -

4 : 3

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

In DC :- In AC:-

i₁ = 4A i₂ = 4 sin $ω t$

R₁ = 3 $Ω$ R₂ = $2 Ω$

In same time internal of ‘t’ - $\frac{H_{1}}{H_{2}} = \frac{i_{1}^{2} R_{1} t}{{(l_{2^{- r m s}})}^{2} R_{2} t} = \frac{16 \times 3 t}{{(\frac{4}{\sqrt[]{2}})}^{2} \times 2 \times t} = \frac{48}{(\frac{32}{2})} = 3 : 1$

In DC :-                In AC:-              i1 = 4A                 i2 = 4 sin <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815297043"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mi>ω</mi> <mi>t</mi> </mrow> </math>               R1 = 3 <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815297044"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mi>Ω</mi> </mrow> </math>             R2 = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815297045"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mn>2</mn> <mi>Ω</mi> </mrow> </math>               In same time internal of ‘t’ - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <msubsup> <mrow> <mi>i</mi> </mrow> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mi>t</mi> </mrow> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi mathvariant="script">l</mi> </mrow> <mrow> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mo>−</mo> <mi>r</mi> <mi>m</mi> <mi>s</mi> </mrow> </msup> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mi>t</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>6</mn> <mo>×</mo> <mn>3</mn> <mi>t</mi> </mrow> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>×</mo> <mn>2</mn> <mo>×</mo> <mi>t</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <mn>8</mn> </mrow> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>3</mn> <mn>2</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mo>:</mo> <mn>1</mn> </mrow> </math>

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A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of $3 Ω a n d 2 Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be:

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A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of 3 Ω a n d 2 Ω respectively. The ratio of heat produced in the two resistances in same interval of time will be:

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A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of $3 Ω a n d 2 Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be: