AP EAMCET · PHYSICS · Capacitance
Two parallel plate capacitors \(8 \mu \mathrm{F}\) each are connected in parallel to a \(10 \mathrm{~V}\) battery. The plate separation in one of the capacitor is reduced to \(40 \%\) of its initial value. The increase in the total charge stored on the capacitors is
- A \(80 \mu \mathrm{C}\)
- B \(120 \mu \mathrm{C}\)
- C \(100 \mu \mathrm{C}\)
- D \(\frac{160}{3} \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(\frac{160}{3} \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Given, \[ \begin{aligned} C_1 & =C_2=8 \mu \mathrm{F}=8 \times 10^{-6} \mathrm{~F} \\ V & =10 \mathrm{~V} \end{aligned} \] Since, \(C_1\) and \(C_2\) are connected in parallel. Equivalent capacitance,…
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