After Current Pulse
Since the switch is closed only for 1 ms i.e. T = 10-3 and opened after the specified time and remains open for E seconds, the amplitude of supplied current drops to zero i.e. I = 0. Since the circuit is not complete, there is no path for the charges stored inside the capacitor to escape and hence the potential difference is maintained inside the capacitor i.e. V0 = 1. Hence voltage across the capacitor is 1 volt. It can be mathematically proven as,
[Tex]V = \frac{1}{C} \int_{T}^{E} Idt+V_{0}\\ \hspace{1mm}\\ V = \frac{1}{{10}^{-6}} \int_{{10}^{-3}}^{E} 0dt+1\\ \hspace{1mm}\\ V = {10}^{6}[({0}\times{E})-({0}\times{{10}^{-3}})]+1[/Tex]
∴ V = 1 volt, for any value of E
Capacitor i-v equation in action
A capacitor is a passive electronic component that can store energy and release it quickly when required. This energy is stored in the form of the electric field which gets created due to the accumulation of electrons on one of the plates of the capacitor. The amount of charge a capacitor can store is called as capacitance and it is measured in Farads named after the scientist Michael Faraday.
A capacitor can be polarized (the anode and cathode are strictly marked and cannot be interchanged) such as an electrolytic capacitor or non-polarized (anode and cathode are swappable) like a ceramic capacitor. In this article we will study the derivation of the capacitor’s i-v equation, voltage response to a current pulse, charging and discharging of the capacitor, and its applications. Let’s begin with the topic.
Table of Content
- Derivation
- Voltage Response to Current Pulse
- Charging of a Capacitor
- Discharging of a capacitor
- Energy Storage and Release
- Applications of Capacitor