AP EAMCET · PHYSICS · Capacitance
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 \(200 \mathrm{~V}\)
- B \(300 \mathrm{~V}\)
- C \(400 \mathrm{~V}\)
- D \(500 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(300 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
\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 }}…
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