Consider the combination of 2 capacitors


Consider the combination of 2 capacitors $C_{1}$ and $C_{2}$, with $C_{2}>C_{1}$, 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, $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$

Note: NTA has dropped this question in the final official answer key.

  1. (1) $\frac{15}{11}$

  2. (2) $\frac{29}{15}$

  3. (3) $\frac{15}{4}$

  4. (4) None of the above

Correct Option: , 4



$C_{1}+C_{2}=\frac{15}{4}\left(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\right)$

$4\left(C_{1}+C_{2}\right)^{2}=15 C_{1} C_{2}$

$4 C_{1}^{2}+4 C_{2}^{2}-7 C_{1} C_{2}=0$

$4+4\left(\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\right)^{2}-7 \frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=0$

$4\left(\frac{C_{2}}{C_{1}}\right)^{2}-7 \frac{C_{2}}{C_{1}}+4=0$

$\frac{C_{2}}{\mathrm{C}_{1}}$ has not real value

$\frac{C_{2}}{C_{1}}=$ imaginary

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now