Question
Download Solution PDFThe energy stored in the capacitor C1 under steady state condition is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option 4):(250 J)
Concept:
The energy stored in the capacitor is
\(\Rightarrow U = \frac{1}{2}C{V^2}\)
Calculation:
Given the circuit is
The voltage across the capacitor is 5V
The energy stored in the capacitor is
U = \(1\over 2\) CV2
= \(1\over 2\) × 20 × 52
= 250 J
Last updated on Jun 16, 2025
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