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Roger V.
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ThinkAnalogy
Think of water running through a pipework: it may run faster where the pipes are narrower, and slower in the wider parts, but the quantity of water which enters is the same as the quantity of water that exits. There may be temporary decompressions here and there, but the water eventually has to come out, unless there is a leak or a reservoir somewhere.

Without analogies
All the charge entering the electric circuit (usually in the form of electrons) should eventually exit it. In narrower wires charge may run faster, but through a smaller cross section, in wider wires it may run slower, but the same quantity of charge will pass through every cross-section... unless there is charge accumulation or a leak somewhere. There may be temporary charge accumulations here and there, but they eventually dissipate.

Note however, that the current density (i.e. charge passing per unit area) will not be the same in the cross-sections of different size.

Mathematically this is described by a so-called continuity equation: $$\frac{dQ}{dt}=I,$$ where $I$ is the current entering the circuit, and $Q$ is the charge (or the quantity of water) accumulated within the circuit. More precise form of this equation formulated for charge and current densities ($\rho$ and $\vec{j}$) is $$\frac{d\rho}{dt} = \nabla\cdot \vec{j},$$ but this might be too mathy.

Think of water running through a pipework: it may run faster where the pipes are narrower, and slower in the wider parts, but the quantity of water which enters is the same as the quantity of water that exits. There may be temporary decompressions here and there, but the water eventually has to come out, unless there is a leak or a reservoir somewhere.

Mathematically this is described by a so-called continuity equation: $$\frac{dQ}{dt}=I,$$ where $I$ is the current entering the circuit, and $Q$ is the charge (or the quantity of water) accumulated within the circuit. More precise form of this equation is $$\frac{d\rho}{dt} = \nabla\cdot \vec{j},$$ but this might be too mathy.

Analogy
Think of water running through a pipework: it may run faster where the pipes are narrower, and slower in the wider parts, but the quantity of water which enters is the same as the quantity of water that exits. There may be temporary decompressions here and there, but the water eventually has to come out, unless there is a leak or a reservoir somewhere.

Without analogies
All the charge entering the electric circuit (usually in the form of electrons) should eventually exit it. In narrower wires charge may run faster, but through a smaller cross section, in wider wires it may run slower, but the same quantity of charge will pass through every cross-section... unless there is charge accumulation or a leak somewhere. There may be temporary charge accumulations here and there, but they eventually dissipate.

Note however, that the current density (i.e. charge passing per unit area) will not be the same in the cross-sections of different size.

Mathematically this is described by a so-called continuity equation: $$\frac{dQ}{dt}=I,$$ where $I$ is the current entering the circuit, and $Q$ is the charge (or the quantity of water) accumulated within the circuit. More precise form of this equation formulated for charge and current densities ($\rho$ and $\vec{j}$) is $$\frac{d\rho}{dt} = \nabla\cdot \vec{j},$$ but this might be too mathy.

Source Link
Roger V.
  • 65k
  • 7
  • 69
  • 215

Think of water running through a pipework: it may run faster where the pipes are narrower, and slower in the wider parts, but the quantity of water which enters is the same as the quantity of water that exits. There may be temporary decompressions here and there, but the water eventually has to come out, unless there is a leak or a reservoir somewhere.

Mathematically this is described by a so-called continuity equation: $$\frac{dQ}{dt}=I,$$ where $I$ is the current entering the circuit, and $Q$ is the charge (or the quantity of water) accumulated within the circuit. More precise form of this equation is $$\frac{d\rho}{dt} = \nabla\cdot \vec{j},$$ but this might be too mathy.