You have two capacitors (call them capacitors A and B). Capacitor A has a capaci

Question

You have two capacitors (call them capacitors A and B). Capacitor A has a capacitance of 2.0 F and capacitor B has a capacitance of 1.0 F. Capacitor B is initially uncharged. Capacitor A is charged by connecting it to a 2.0 V battery. Capacitor A is then disconnected from the battery and connected to Capacitor B (as shown in the figure to the right). What electrical potential energy is stored in the two capacitors after they have been connected together? a. 2.0 J. b. 1.3 J. c. 4.0 J. d. 2.7 J. e. 1.0 J.

Explanation / Answer

The charge stored by the capacitor A is

QA = CA V = 2.0 F * 2.0 V = 4.0 C

this is the total charge of the system forms by the two capacitors

QT =  4.0 C

since the capacitors are in parallel their equivalent capactance is

Ceq = CA + CB = 2.0 F + 1.0 F = 3.0 F

hence, the energy of the system is

U = QT2 / ( 2 Ceq )

U = ( 4.0 C )2 / ( 2 * 3.0 F )

U = 2.7 J

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