We appreciate your visit to A 20 F capacitor is charged to 5 V and isolated It is then connected in parallel with an uncharged capacitor of 30 F The. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
The decrease in energy when a 20F capacitor charged to 5V is connected in parallel with an uncharged 30F capacitor is 150J (D).
The question inquires about the change in energy when a fully charged 20F capacitor is connected in parallel with an uncharged 30F capacitor. Initially, the charged capacitor has an energy of 0.5 x 20F x (5V)^2 = 250J.
When the capacitors are connected in parallel, the total capacitance becomes 20F + 30F = 50F, and the charge remains the same, distributing over the larger total capacitance.
The final voltage V across both capacitors is given by the initial charge Q (which is 20F x 5V = 100C) over the total capacitance C, which is V = Q/C = 100C/50F = 2V.
The final energy of the system is 0.5 x 50F x (2V)^2 = 100J.
Therefore, the decrease in energy is 250J - 100J = 150J.
Thanks for taking the time to read A 20 F capacitor is charged to 5 V and isolated It is then connected in parallel with an uncharged capacitor of 30 F The. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
