Consider the combination of 2 capacitors C1 and C2, with C2 > C1, when connected in parallel, the equivalent capacitance is <math><mrow><mfrac><mrow><mn>1</mn><mn>5</mn></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></math> times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors, <math><mrow><mfrac><mrow><msub><mrow><mi>C</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow><mrow><msub><mrow><mi>C</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></mfrac><mo>.</mo></mrow></math>

According question, we can writeC1×C2C1+C2=154 (C1+C2)⇒4C1C2=15 (C12+C22+2C1C2)⇒4 (C2C1)=15 (C2C1)2+30 (C2C1)+15⇒15x2+26x+15=0, where x = C2C1⇒x=−26±676−90030Since x cannot be real, so this Question has been dropped by NTA

Consider the combination of 2 capacitors C1 and C2, with C2 &gt; C1, when connected in parallel, the equivalent capacitance is&nbsp;<math><mrow><mfrac><mrow><mn>1</mn><mn>5</mn></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></math>&nbsp;times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors,&nbsp;<math><mrow><mfrac><mrow><msub><mrow><mi>C</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow><mrow><msub><mrow><mi>C</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></mfrac><mo>.</mo></mrow></math>

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Physics Class 12th Physics Electrostatic Potential and Capacitance

Consider the combination of 2 capacitors C₁ and C₂, with C₂ > C₁, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors, $\frac{C_{2}}{C_{1}} .$

Option 1 -

$\frac{111}{80}$

Option 2 -

$\frac{15}{4}$

Option 3 -

$\frac{29}{15}$

Option 4 -

$\frac{15}{11}$

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

According question, we can write

$\frac{C_{1} \times C_{2}}{C_{1} + C_{2}} = \frac{15}{4} (C_{1} + C_{2}) \Rightarrow 4 C_{1} C_{2} = 15 (C_{1}^{2} + C_{2}^{2} + 2 C_{1} C_{2})$

$\Rightarrow 4 (\frac{C_{2}}{C_{1}}) = 15 {(\frac{C_{2}}{C_{1}})}^{2} + 30 (\frac{C_{2}}{C_{1}}) + 15 \Rightarrow 15 x^{2} + 26 x + 15 = 0,$ where x = $\frac{C_{2}}{C_{1}}$

$\Rightarrow x = \frac{- 26 \pm \sqrt[]{676 - 900}}{30}$

Since x cannot be real, so this Question has been dropped by NTA

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Consider the combination of 2 capacitors C₁ and C₂, with C₂ > C₁, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors, $\frac{C_{2}}{C_{1}} .$

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Consider the combination of 2 capacitors C1 and C2, with C2 > C1, when connected in parallel, the equivalent capacitance is 154 times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors, C2C1.

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Consider the combination of 2 capacitors C₁ and C₂, with C₂ > C₁, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors, $\frac{C_{2}}{C_{1}} .$