How Do Capacitors Behave and Function in Electrical Circuits?

[OJ-Hick]

This is not a homework question for the record. An external question which came up in my revision!

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http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 2000/June 2007/6735_01_que_20070614.pdf

QUESTION 3

Parts B and C

 

[Stonebridge]

What do you think would happen?
Have you given it any thought?

 

[kerimek]

of this question involve a circuit containing a capacitor that is initially uncharged.

Part B asks for the energy stored in the capacitor after it has been charged for 2 seconds.

Part C asks for the current through the capacitor after it has been charged for 2 seconds.

I would approach this problem by first understanding the concept of a capacitor and its behavior in an electrical circuit. A capacitor is a device that stores electrical energy in the form of an electric field. When connected to a power source, it charges up and stores energy, and when disconnected, it can release this stored energy.

To calculate the energy stored in the capacitor after 2 seconds, I would use the equation E = 1/2 * C * V^2, where E is the energy, C is the capacitance, and V is the voltage across the capacitor. The voltage across the capacitor can be found using the equation V = Q/C, where Q is the charge on the capacitor. Since the circuit is initially uncharged, the charge on the capacitor is 0, and therefore the voltage is also 0. This means that the energy stored in the capacitor after 2 seconds is also 0.

To calculate the current through the capacitor after 2 seconds, I would use the equation I = C * dV/dt, where I is the current, C is the capacitance, and dV/dt is the rate of change of voltage with respect to time. In this case, since the capacitor is already fully charged after 2 seconds, there is no change in voltage, and therefore the current through the capacitor is also 0.

In conclusion, the energy stored in the capacitor and the current through the capacitor after 2 seconds are both 0 in this scenario. This is due to the fact that the capacitor is already fully charged and there is no further change in voltage or current. It is important to understand the behavior of capacitors in electrical circuits in order to accurately solve problems and understand their role in various applications.