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If electric current is the time rate of CHANGE of charge. Then if I have a DC fixed current flowing into a resistor, same amount of charge is always flowing to that resistor, hence the derivative of charge is 0, so current is 0?

Or does the book mean that the electric current is the time rate of change of charge TRANSFERRED - in that case, the total q transferred will always be increasing, hence it is possible to have a fixed current?

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  • \$\begingroup\$ my understanding is as you said on second statement, that current is the time rate of change of charge transferred. could be between two points on a circuit, or cables, etc. if the total q always changed, doesn't the total time will change too? this is still my understanding and everyone is free to correct me. thank you \$\endgroup\$
    – Rinn D
    Apr 21 at 2:28
  • \$\begingroup\$ Think the second definition but more the derivative of "total number of charges that have passed through this point" being the current. \$\endgroup\$
    – DKNguyen
    Apr 21 at 2:36
  • \$\begingroup\$ That would make more sense to me as well. The definition in the book makes it seem like they are talking about the derivative of the instantaneous number of charges flowing, which would always be a constant for a resistive circuit. \$\endgroup\$ Apr 21 at 3:18
  • \$\begingroup\$ "Then if I have a DC fixed current flowing into a resistor, same amount of charge is always flowing to that resistor, hence the derivative of charge is 0, so current is 0?" -- no, you took one derivative too many. The time derivative of charge is the flow of charge, or current. The time derivative of that is the change in current over time, which for DC is zero. \$\endgroup\$
    – ilkkachu
    Apr 21 at 10:56
  • \$\begingroup\$ Technically, an electrically powered device is USING power meaning it is depleting power. Keep that in mind as a fundamental basis of your understanding. I would recommend to try to keep a practical understanding, don't lose the forest for the trees. \$\endgroup\$ Apr 21 at 18:09

4 Answers 4

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Current flows through things. Even a capacitor that stores up charge will have the same amount of current coming out of one terminal that's going into the other.

So the current through a wire is defined as the amount of charge that passes by a point in the wire at a given time.

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    \$\begingroup\$ To clarify, let's say I took an theoretical invisible wire mesh and placed in across a wire, the current would be the derivative of the TOTAL number of charges that pass through that wire. If the device on the other end is receiving a fixed DC current, then the total number of charges that pass through my mesh will be increasing linearly, hence the derivative being a constant current. Would that be correct? Really my confusion is with the fact that it looks like they are talking about instantaneous charge passing, rather than the total. \$\endgroup\$ Apr 21 at 3:21
  • \$\begingroup\$ Yes, that's correct. Current is the rate of flow (or derivative) of total charge. \$\endgroup\$
    – TimWescott
    Apr 21 at 16:15
  • \$\begingroup\$ Ok, so just to be sure. the q in dq/dt is talking about the total charge that ever passed through some point in the circuit. Theoretically, we can measure this q by putting an imaginary box in the circuit that just counts all the positive charges that pass through it. The q would be increasing every time a positive charge passes through our box. If we get the derivative of this q, that would be the instantaneous current. For a DC circuit, the rate that the q increases by is a constant, hence the current is constant. \$\endgroup\$ Apr 24 at 2:27
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The equation mentioned in the book seems to be in the context of a battery, I think. Here, the "time rate of change of charge" is talking about the "charge of the battery". From this perspective, relation (1.1) is certainly true. But time rate of change should not be taken too literally: the book mentions that "current is the net flow of positive charges", but there obviously is no net flow of positive charges into a resistor and a resistor does not accumulate charge, so also dQ/dt=0. That would be a contradiction.

From a more general point of view, charge can flow through something, like a resistor or a wire, or into something, like a capacitor or a battery. In the first case, you would measure the "electrons passing by", in the second the "electrons coming in/out" (per time).

In both cases we can legitimately call it "current" because their unit is Ampere:

  • if we say, current is the transfer of charge per time we get Q/t [C/s = As/s = A]
  • if current is the time rate of change of charge we also get dQ/dt [C/s = As/s = A]
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If electric current is the time rate of CHANGE of charge. Then if I have a DC fixed current flowing into a resistor, same amount of charge is always flowing to that resistor, hence the derivative of charge is 0, so current is 0?

Current is typically measured as either the amount of charge flowing past a point on a 1D wire in simple lumped circuit models, or charge flowing through a surface in more sophisticated models.

The current into each terminal of the resistor is not zero. But the current into one leg is going to have opposite sign of the current going out of the other leg. So, the sum of the two currents would be zero.

This is analogous to Kirkhoff's current law where the sum of the currents into/out-of any node is zero.

On the other hand, if the net current into a region of space is not zero, then charge would be building up and we would have an electric field according to Gaus's law

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  • \$\begingroup\$ About the Gaus's law point - how does that apply to a capacitor? A +Q charge on a capacitor top plate will be mirrored by a -Q charge on the bottom plate? Does that mean a capacitor does not have an electric field since the net charge is 0. \$\endgroup\$ Apr 23 at 18:10
  • \$\begingroup\$ @alayoiskgfbfqhxjiw I was not really intending to make a connection to a capacitor with that point. But Each plate on a capacitor viewed alone certainly has a net charge on it. There is in fact an electric field in the region between the plates. But viewed from distances that are far away compared to the size of the physical features of the capacitor the opposite fields on each plate appear to cancel out. If one had an enormous air-gap capacitor charged to very high voltage they could certainly walk around through it with a test charge and measure the field. \$\endgroup\$
    – user4574
    Apr 23 at 22:57
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There is no error in the excerpt in question.

Current flows thru the resistor. Ideal conductors remain neutral. However, there will be charges in the insulating medium attracted and surrounding the conductors from the Electric field.

The current would simply be measured across resistance as I=V/R.

Since all conducting wires also have a dielectric which includes air , two isolated wires in close proximity will always have a capacitance between them and can transfer charges when the electric field is rapidly changing Ic = C*dV/dt. All wires also have inductance which can store some insulation energy as current from the square of the charge carrier flow rate Ec=½LI².

Thus for high frequency AC, all wires will have some impedance due to this ratio of conductor current and insulation current and can be expressed in terms of the reactive lumped values Zo=√(L/C).

The average velocity of a charged particles is NOT current. That's called Drift Velocity.

The velocity of the wave of charged particles is NOT current. That's called the wave propagation velocity, which is equal to the speed of light.

"An electric current [Amps] is a stream of charged carriers , such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume" (using volume resistivity). However, conventionally using lumped elements current is measured by the voltage drop across a known precise very small resistance called a current sensing shunt resistor.

There is a vast number of free charges in conductors that can be excited to move from a voltage source across a closed circuit loop with low resistance. I=V/R Voltage is just the potential to move charges. Current can flow in insulators according to an exponential rate and their capacitance C. These charges are then stored. Q=CV by a rate of change of voltage Ic = C dV/dt.

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