Find the energy stored in each capacitor and inductor
Answers
Answer:
The circuit contains only dc sources. Recall that an inductor is a short circuit to dc and a capacitor is an open circuit to dc. These can be easily verified from their current-voltage characteristics. For an inductor, we have v(t)=Ldi(t)dt. Since a dc current does not vary with time, di(t)dt=0. Hence, the voltage across the inductor is zero for any dc current. This is to say that dc current passes through the inductor without any voltage drop, exactly similar to a short circuit. For a capacitor, the current-voltage terminal characteristics is i(t)=Ldv(t)dt. Voltage drop across passive elements due to dc currents does not vary with time. Therefore, dv(t)dt=0 and consequently the current of the capacitor is zero. This is to say that dc current does not pass through the capacitor regardless of the voltage amount. This is similar to the behavior of an open circuit. Please note that unlike dc current, ac current passes through capacitors in general.
To find dc current of the inductors and dc voltage drop across the capacitors, we replace them with their equivalent elements, short circuits and open circuits, respectively, as shown in Fig. (1-28-2).
Energy stored in circuit - dc equivalent
Fig. (1-28-2) - Replacing inductors and capacitors with their dc equivalents
It is easy to find that I1Ω=9V2Ω+1Ω=3A and I3mH=1A. Therefore, I2mH=I3mH+I1Ω=4A, V20μF=−1Ω×I1Ω=−3V and V10μF=−2Ω×I1Ω=−6V. The total energy stored in the circuit is the sum of the energy stored in elements capable of storing energy, i.e. two capacitors and two inductors. Recall that the energy stored in an inductor is wL=12Li2L(t) and is equal to wC=12CV2C(t) for a capacitor. Thus,
w3mH=12(3mH)(1A)2=1.5mJ
w2mH=12(2mH)(4A)2=16mJ