Q146 / 160 · AP EAMCET 2023 (15 May Shift 2) · Physics

Two capacitors of capacity $4 \mu \mathrm{F}$ and $6 \mu \mathrm{F}$ are connected in series to a $500 \mathrm{~V}$ battery. The potential difference across $4 \mu \mathrm{F}$ capacitor is

  • A
    $200 \mathrm{~V}$
  • Correct answer
    B
    $300 \mathrm{~V}$
  • C
    $400 \mathrm{~V}$
  • D
    $500 \mathrm{~V}$

Solution

$\begin{aligned} & \mathrm{C}_1=4 \mu \mathrm{F} \text { and } \mathrm{C}_2=6 \mu \mathrm{F} \\ & \mathrm{C}_{\text {eq }}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2} \\ & =\frac{4 \times 6}{4+6}=2.4 \mu \mathrm{F} \\ & \mathrm{Q}=\mathrm{C}_{\text {eq }} \mathrm{V} \\ & =2.4 \times 10^{-6} \times 500=1200 \times 10^{-6} \mathrm{C} \end{aligned}$ The potential difference across $4 \mu \mathrm{F}$ capacitor is $\begin{aligned} & \Delta \mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}_1} \\ & =\frac{1200 \times 10^{-6}}{4 \times 10^{-6}}=300 \mathrm{~V} \end{aligned}$

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

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