AP EAMCET · PHYSICS · Capacitance
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 \(1000 \mathrm{~V}\)
- 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 \(2500 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(10^{-6} \mathrm{~F}\) and \(10000 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given, capacitance \(C=2 \times 10^{-6} \mathrm{~F}\) Potential difference break down voltage, \[ V=5000 \mathrm{~V} \] Capacitance in series, \(\frac{1}{C_S}=\frac{1}{C}+\frac{1}{C}\)…
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