Saturday, January 10, 2026

Charging of capacitor whole process from start to finish

 The charging of a capacitor is the process of storing electrical energy in an electrostatic field by transferring charge onto its plates. When an uncharged capacitor ($C$) is connected in series with a resistor ($R$) and a DC voltage source ($V_s$), the charging process is not instant but exponential, governed by the time constant ($\tau = RC$). [1, 2, 3]


1. Initial State ($t = 0$)
  • Conditions: The capacitor is completely discharged. Voltage across the capacitor ($V_c$) is 0V, and charge ($q$) is 0.
  • Behavior: When the switch is closed, the capacitor acts like a short circuit (a simple wire).
  • Current: The maximum current flows, determined only by the resistor: $I_{max} = V_s / R$. [2, 4, 5, 6, 7]
2. Charging Phase ($0 < t < 5\tau$)
  • Mechanism: Electrons are pulled from the plate connected to the positive terminal of the battery and pushed onto the plate connected to the negative terminal.
  • Voltage Growth: As charge ($q$) accumulates, the voltage across the capacitor ($V_c$) increases exponentially, according to the formula:$V_c(t) = V_s(1 - e^{-t/RC})$
  • Current Decay: As $V_c$ approaches $V_s$, the net voltage pushing current ($V_s - V_c$) drops. The current decreases exponentially:$I(t) = (V_s/R)e^{-t/RC}$
  • Time Constant (): The product $RC$ (in seconds) defines how fast the capacitor charges. After one time constant ($t = RC$), the capacitor reaches ~63.2% of its maximum charge. [1, 5, 8, 9, 10]
3. Final State (Steady State: $t > 5\tau$)
  • Conditions: After approximately 5 time constants ($5\tau$), the capacitor is considered fully charged.
  • Voltage: $V_c$ is equal to the supply voltage $V_s$.
  • Current: No more charge can accumulate, so current ($I$) becomes zero.
  • Behavior: The capacitor acts as an open circuit (breaking the circuit). [1, 11, 12, 13]
4. Summary Table of Charging
Parameter [14, 15, 16, 17, 18] Start ($t=0$) During Final ($t>5\tau$)
Voltage ($V_c$) 0 Increases ($1-e^{-t/RC}$) $V_s$
Current ($I$) $V_s/R$ (Max) Decreases ($e^{-t/RC}$) 0
Charge ($q$) 0 Increases $CV_s$ (Max)
Component Short Circuit - Open Circuit

Key Takeaways
  • Energy Storage: The capacitor stores energy as a potential difference between its plates, equal to $E = \frac{1}{2}CV_s^2$.
  • Role of R and C: A larger resistance ($R$) or larger capacitance ($C$) slows down the charging process (increases the time constant).
  • Energy Loss: During the charging process, half of the total energy supplied by the source is dissipated in the resistor, while only half is stored in the capacitor. [1, 11, 19, 20]

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