energy stored in circuit under steady circuit

Hello there i want to solve this problem

I solved above problem this way. In steady state condition, capacitor should be replaced by open circuit. so 2 ohm, 4 ohm and 2 ohm these three resistors are in series. so total resistor is 8 ohm. and resistor only dissipate energy. so energy stored in this circuit should be zero.

but correct answer is 288 uJ. can anyone tell me?? why i am wrong??

• In the steady state, the capacitors are charged (depending on the voltage drop over them).
– CL.
Mar 10, 2017 at 11:37
• can you help me to solve it?? Mar 10, 2017 at 11:49
• Do you know how to compute the energy in a capacitor with a known voltage?
– CL.
Mar 10, 2017 at 11:49
• In the steady state, the frequency is zero. You are overthinking this; you already know the voltages at A,B,C,D in the steady state.
– CL.
Mar 10, 2017 at 11:57
• Technically, the battery is a part of the circuit in this question, so the stored energy will be primarily defined by its capacity. Mar 10, 2017 at 12:32

The energy stored in the circuit in steady-state is not zero even though the frequency is zero, because the capacitors will be charged. As you correctly stated in your question you can remove the caps to determine the voltage drops across the resistors. You are also correct that only the resistors will dissipate energy, but the caps will stay charged, therefore storing energy.

Once you determine the voltage across the resistors, you can put the caps back in and assume they will be charged to the same level as the voltage across the respective connection points. Then use the energy formula E = 0.5CV^2.

• thanks. because of capacitor i am confusing in this question. if i replace capacitor with resistor than i can write Vad = (2/(2+2))*16. but what should i write in case of capacitor?? Mar 10, 2017 at 12:12
• No, don't replace the caps with resistors, just remove the caps as you said, then find all the voltages. Where did you get those values (2/(2+2)) ? Vad would be 16 * Rad / (Rad + Rdc) = 16 * (Rab + Rbd) / (Rab + Rbd + Rdc) = 16 * (2 + 4) / (2 + 4 + 2) = 16 * 6 / 8 Mar 10, 2017 at 12:27
• ok sorry its my mistake. so vad = 12 v and vbc = 12 v ?? Mar 10, 2017 at 12:36
• Yes, correct, the caps are both at 12V, and the energy calculation looks right. But it doesn't match the 1.48uJ you specified. Are you sure that's the correct answer? Mar 10, 2017 at 13:18
• actually i made mistake and i edited this question Mar 10, 2017 at 13:22

Find voltage between nodes A and D and between nodes B and C ("steady state" means assume long enough charging time; i.e. ignore capacitors for finding voltage).

Energy stored in one capacitor is E=1/2 CV².

• thanks. because of capacitor i am confusing in this question. if i replace capacitor with resistor than i can write Vad = (2/(2+2))*16. but what should i write in case of capacitor?? Mar 10, 2017 at 12:10
• i think Vad= Vbc = 16 v. so total energy is 512uJ. but it didn't match with my answer Mar 10, 2017 at 12:20
• @Beginner: as I said: ignore capacitors for finding voltage!
– Curd
Mar 10, 2017 at 12:58
• ok i understand my mistake and i got solution Mar 10, 2017 at 12:59

in easy way u can get friends just understand that in steady state caps are open circuited so total resistence is 8 ohm so if we calculate the total current i =2amp so individual voltage drops are 4v,8v,4v for 2ohm,4ohm,2ohm respectively so open curcuit means there will be voltage so calculate now 16v incoming 4v going to 2ohms so remaing 12v into cap and from 12v .8v to 4ohm,then remaining 4v to 3ohm so the cap are having 12v each. so by applying E=0.5 cv2 we get 144uj each so if we add 288uj so answer is 2.88x10^-4

• -1 for lack of capital letters and missing punctuation. Very messy and difficult to read. Oct 4, 2017 at 12:18