Question
Download Solution PDFA capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Capacitor is defined as a component in an electrical circuit which has the ability to store energy in the form of an electrical charge producing a potential difference across its plates. It is like rechargeable battery.
Charge on capacitor is given by:
q = CV
where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.
Energy stored in capacitor is calculated using the formula, U:
\(\rm U= \frac{1}{2}C V^2\)
where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.
Calculation:
Charge on capacitor
q = CV
When it is connected with another uncharged capacitor
\(V_c =\frac{q_1+q_2}{C_1 + C_2}= \frac{q + 0}{C + C}\)
\(\rm V_c= \frac{V}{2}\)
Initial energy
\(\rm U_f = \frac{1}{2}C \left(\frac{V}{2}\right)^2 + \)\(\rm\frac{1}{2}C\left(\frac{V}{2}\right)^2\)
\(= \rm\frac{CV^2}{4}\)
Loss of energy = Ui - Uf
\(\rm=\frac{CV^2}{4}\)
i.e. decreases by a factor (2)
Last updated on Jun 16, 2025
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