Before the switch is closed in a circuit consisting of a 4μC capacitor and 0.2 H inductor, the potential across the capacitor is 200 V. At some instant after the switch is closed, the instantaneous current is 0.7 A. What is the energy in the capacitor at this instant?Answer: 31.0 mJPlease show steps. Thank you.
Power in watts equals volts x amps
The total energy before the switch is closed is: 1/2CV^2 2e-6*40000 0.08J 80 mJ At the moment when I 0.7A, the energy in the inductor is 1/2*L*I^2 0.1*0.7^2 0.049J 49 mJ The rest of the energy must be in the capacitor 80-4931mJ Notice that there has to be some resistance in the circuit which would gradually consume the energy. The inductor is modeled as having no resistance but that is ideal.
