I have confusion regarding the total power cnsumed by the series Rc circuit when connected with the DC pulse current source.
the power calculated for this cicuit is I^(2)*R ?? Is there any power loss in capacitor?

 

No, ideal capacitors and inductors have no average power, however they can charge and discharge to provide instantaneous power (so only temporarily).
So, power into a capacitor isn't lost, it's being stored and you can use it later.

 

thankyou for your reply. so it means for calculating the power consumption over time ( energy consumed )in this circuit i can ignore the capacitor and as current is contant and i know the resistance value I can use the I^2*R and integrate this value over time?

 

If you only want power loss then you only need to look at the resistor.
However, the current will not be constant, because even for a DC source it will be changing for a while until you hit a steady state.
So unless you're only interested in the steady state you will need to solve the circuit anyway, and then power comes easily from knowing the voltages and currents in the circuit.

But yes, if you're looking at the steady state with constant current in the resistor you can just integrate that.

 

the source is pulse current source. In the circuit when the pulse is high the capacitor is charging through resistor (the current remains constant at high value suplied by current source) and when its low that is current is zero( by the source) the capacitor discharges so there is a change of volatge across the capacitor but no current in the circuit .
So i want to know that i only need to calculate power consumed when the current cycle is high? because when current is zero power is zero even when the capacitor is discharging ( that is the part i m confuse about)??

 

If it is a current source in series with a resistor and capacitor then once current is zero the capacitor does not discharge.
A 0 amp current source is an open circuit.

However, I now think you're being asked for the power consumption from the source, in which case you do need to figure out how much energy is being stored by the capacitor.

If the capacitor was discharging there is power coming from the capacitor. Average power is taken over all time, so capacitors have zero average power because to extract 1 Joule you first need to store 1 Joule. In your circuit you are storing energy in the capacitor but not extracting it.

 

yes you r right I need to calculate the power consumption from the source.
So now i know the resitance value , current value , delta V across capacitor ( while capacitor is charging) and capacitance
so for energy calculation in one cycle
can I use the following equation?
(I^2 *R )*high pulse time + 1/2 (delta V)^2 * C

and mutiply the above value with frequency to get the energy consumed from source in one second

just to give you more detail of my system i have attached the volatge aand current waveform ( I have calculated all the parameters mentioned above from these wavefroms)
Thankyou for your help

 

That calculation should work, but only because your initial voltage is 0.
The change in stored energy is:
\(\frac{1}{2}C(V_2^2 - V_1^2)
\)
You have V1 = 0

It's interesting because your capacitor isn't discharging, at least in your waveform and your description.
This means V1 won't be 0 anymore for the second pulse.
The voltage will increase with every pulse until the current source can no longer work properly.
Since that capacitor voltage keeps rising it will actually absorb more energy from each pulse.

 

no it does discharge , before the next pulse it will be again zero but it is not shown in that figure( its just one pulse magnified). Capacitor is discharging very slowly and before the next pulse the value is agian zero like the figure i have shown. Thats why i was askin that can I use the calcualtion for one pulse and multiply with the frequency to get the power consumption in one second.
Thankyou so much for your help

 

No problem