In this source-R-L series circuit (I redraw a circuit with different parameters attached below), assuming v1(t) for the instantaneous voltage across the source, vR(t) for the voltage across the resistor, vL(t) for the voltage across the inductor. By KVL, we haveAt some points the source will also absorb energy, and that happens for a short time just after the source voltage goes through zero. The inductor supplies all the energy when the source is at or near zero. When the source absorbs energy it may be considered a loss or a storage, depending on what kind of source it is.
In a circuit in which an AC source, R,L are connected in series, the average/real/active power delivered to combination of R and L is, according to its definition, the average of product of the instantaneous source voltage and the real part of the source current in one period of the instantaneous source voltage/current.
Is it correct that the average power can also be obtained by calculating the average of product of the instantaneous voltage across the RESISTOR and the current through the RESISTOR in one period?
So I thought that when the instantaneous power supplied by the source [i(t)*v1(t)] is positive, the power absorbed by the inductor [i(t)*vL(t)] is negative (indicating that the inductor is supplying energy), then the instantaneous power dissipated by the resistor is supplied by BOTH the voltage source and the inductor, is this correct?
But when i(t)*v1(t) and i(t)*vL(t) are negative, and | i(t)*vL(t) | > | i(t)*v1(t) |, then where does the power | i(t)*vL(t) |- i(t)*vR(t) go? Is it "consumed" by the ideal voltage source?
I used to connect a 3V battery and a 1.5V one in parallel, but I didn't feel any noticeable warm on the connecting wire. Is this connection a complete circuit? How is the electric energy consumed?
I would like to focus on the period (between 10ms ~ 14ms in Voltages&Current.jpg below) where the instantaneous voltage across the source V_S(t) is negative (the Green curve in Voltages¤t.jpg below), the voltage across inductor V_L(t) (Red) is negative while | V_L(t) | is greater than | V_S(t) |, and the current (Blue) through the entire circuit i(t) is positive and decreasing, naming
during 10ms < t < 14ms, V_S(t)<0, V_L(t)<0, V_L(t)<V_S(t)
In THIS case, it seems to me that the power delivered/returned by the inductor has tow parts, one part is dissipated by the resistor, that is , [V_L(t) - V_S(t)] * i(t) is dissipated by the resistor, but (1)where does the other part, V_S(t)*i(t), go, in what form(s)?
Assume that the source voltage is NOT an ideal voltage, but is the voltage from an electric company, (2)what influence does this part of reactive power have?
Sorry, the curves were in initial transient state, I should have included the initial condition for the inductor, that is, the initial current = -0.57805A. The instantaneous power curves are also attached, P_S(t) represents power supplied by source, P_L(t) for power absorbed by inductor, and P_R(t) for power dissipated by resistor.All three curves you submitted have different periods. How can that be? Where is the curve representing power? Why don't your curves look like the ones I submitted?
I really cannot understand the description that all the energy in the inductor is stored and returned to the circuit in equal amounts. In this particular RL series circuit with voltage source, ignoring the resistance of the wire, 'returned to the circuit' means returnning to what element? None of the energy stored in inductor is delivered to the resistor?All the energy in the inductor is stored and returned to the circuit in equal amounts.
If I didn't get any calculation errors,None of it is dissipated in the resistor because the phases of the voltages of the inductor and resistor are 90° apart
Ooops! You and Ratch are right. I was wrong.Since the current is 90 degrees out of phase with the voltage in the inductor, that means the resistor current is 90 degrees out of phase with the voltage in the inductor, so that means the voltage across the resistor is 90 degrees out of phase with the inductor voltage.
In the RL series circuit with AC source case, I guess this would hold:But when the line voltage goes to zero (twice per cycle) the inductor still contains energy, and the resistor is dissipating power. Therefore the resistor gets it power from the inductor alone during that brief period, and after which it would get it's power from the inductor and the line.
I guess the point here is that you want me to analyze the circuit in its steady AC state.(?)I think it might help to do two things here:
1. Plot the energy stored in the inductor (not i(v)*vL(L) which is different). The energy is (1/2)*L*i^2 and is always positive.
2. Start with the initial inductor current of zero as before, but this time look at the waveform after 5 time constants, which would be 50ms, but to sync with the line period start looking at 60ms. A good time to view is between 60ms and 80ms. This ensures that all transient phenomena have died out.
3. Plot the resistor power.
4. Try to look at times when something goes through zero. When something goes through zero that eliminates that as a contribution to the circuit because a zero voltage can not deliver nor absorb power for example.
Please forgive me for my poor English. I'm not sure I get the pointIf the line is supplied with energy and you dont have anything else plugged in, then you are not supplying the neighbors with energy
Sorry, I cannot understand this paragraph completely. Could you please tell me more, in terms of easier English? Thank you!This is in fact a key point with solar powered line tied inverters that are used to convert solar energy into electrical energy and pump it into the line. There we do it on purpose, but with the inductor resistor circuit it's just natural.
I can understand why it is so, but I can't understand the following descriptionNote also that if the 1.5v batt had a large series resistance inside of it, the terminal voltage would go up to almost 3 volts.
In what manner a large resistor was inserted into the line/inductor/rsistor circuit would result in a much higher line voltage? Isn't the amplitude of the line voltage fixed?If this happened in the line/inductor/resistor circuit the line voltage could shoot up to a much higher value ruining other equipment plugged into the line.
v
+---R1-----+----R-----+
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Vs1 Vs2
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+---------------------+
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