Q150 / 160 · AP EAMCET 2021 (25 Aug Shift 1) · Physics

Two capacitors each having a capacitance $2 \times 10^{-6} \mathrm{~F}$ and a breakdown voltage $5000 \mathrm{~V}$, are joined in series. What will be the resultant capacitance and the breakdown voltage of the combination?

  • A
    $4 \times 10^{-6} \mathrm{~F}$ and $1000 \mathrm{~V}$
  • Correct answer
    B
    $10^{-6} \mathrm{~F}$ and $10000 \mathrm{~V}$
  • C
    $2 \times 10^{-6} \mathrm{~F}$ and $5000 \mathrm{~V}$
  • D
    $10^{-6} \mathrm{~F}$ and $2500 \mathrm{~V}$

Solution

Given, capacitance $C=2 \times 10^{-6} \mathrm{~F}$ Potential difference break down voltage, $$ V=5000 \mathrm{~V} $$ Capacitance in series, $\frac{1}{C_S}=\frac{1}{C}+\frac{1}{C}$ $$ \begin{aligned} \frac{1}{C_S} & =\left(\frac{1}{2}+\frac{1}{2}\right) \times \frac{1}{10^{-6}} \\ \Rightarrow \quad C_S & =10^{-6} \mathrm{~F} \end{aligned} $$ Potential difference break down voltage in series, $$ \begin{aligned} V_S & =V_1+V_2 \\ & =5000+5000=10000 \mathrm{~V} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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