energy
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??
HINT
In DC steady state, the voltage at the top of the 5V supply and the top of the 4 ohm resistor will be the same. Then your circuit just looks like this:
simulate this circuit – Schematic created using CircuitLab
Use Thevenin's theorem to simplify the voltage divider on the left hand side, and you'll get a circuit containing just a voltage source, a current source, an inductor, and resistor. The current through L1 will be 6A, so you can calculate the energy in that inductor. The same current must flow into the Thevenin equivalent voltage source, so you can easily find the current through the first inductor by equating powers: \$V_{th}*6A = 5*I\$ and solving for I. This will be the current through the first inductor. The voltage across the capacitor in steady state will be 5 volts, so you know the energy in it. Add them up to get the total energy in the circuit.
The 5-second overload rating tells you a couple of things. One, as you note, is an implied "total Joule" rating, which says something about the thermal characteristics of the resistor assembly overall.
The other is that the manufacturer expects that the resistance wire can handle about 700 mA for at least that amount of time. You're asking it to briefly carry twice that current, which may or may not push the resistance wire into a nonlinear region, which could lead to localized failures. With a time constant of 1.65 seconds, it takes about 1.2 seconds for the current to drop from 1.46 A to 700 mA.
I would expect that you'll be fine, but it would be worthwhile to do some tests of your own, running short pulses of current in the range of 1A - 2A through the resistor and monitoring for anomalous resistance changes (i.e., excessive voltage drop).
Or you could just ask the manufacturer directly ...
Best Answer
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.