Meaning of Current in Ohm's law

[FS98]

If we had a parallel circuit with a voltage of V between the beginning and end, and the circuit has a resistance R, then the current given by ohms law is I = V/R.

What does this mean? The current is not the same throughout the whole circuit. Where is the current equal to this value?

 

[FactChecker]

When you say that the "circuit has a resistance R", you must specify two points in the circuit that the resistance is measured between. (Often the two points are considered to be self explanitory and are not specified.) The current from one of those points to the other is the current of Ohm's law.

 

[FS98]

When you say that the "circuit has a resistance R", you must specify two points in the circuit that the resistance is measured between. (Often the two points are considered to be self explanitory and are not specified.) The current from one of those points to the other is the current of Ohm's law.

So if R is the resistance between a point at the beginning and at the end, and V is a voltage between a point at the beginning and at the end, V/R would be the current coming from the beginning and entering the end of the circuit?

It can’t be the current along the entire path between the two points because the current splits at junctions.

 

[russ_watters]

So if R is the resistance between a point at the beginning and at the end, and V is a voltage between a point at the beginning and at the end, V/R would be the current coming from the beginning and entering the end of the circuit?

It can’t be the current along the entire path between the two points because the current splits at junctions.

Er, well, I would say that by conservation of mass the current flow has to be the same *along* the entire path. to me, using the word "along" and singular "path" says you are drawing a line that cuts through both sub-paths and summing their currents to get the total. And you are, conceptually, if not mathematically.

 

[K Murty]

If we had a parallel circuit with a voltage of V between the beginning and end, and the circuit has a resistance R, then the current given by ohms law is I = V/R.

What does this mean? The current is not the same throughout the whole circuit. Where is the current equal to this value?

In a series DC circuit, the current is the same throughout all elements, in a parallel circuit the current differs according to the branch resistance.
I recommend you google Lumped matter discipline and familiarise yourself with general circuit topology such as nodes, branches, loops and meshes, and ideal wires (short circuits) and open circuits, it would work that way to build the understanding IMO. Also the fundamental theorem of network topology. Because in circuit analysis you are now in a specialised domain and it is best to familiarize oneself with the lumped matter discipline and lumped elements and the structure of circuits first.

When you say a parallel circuit of one resistance only, it is the special case of being a circuit which is both series and parallel, one voltage source with one resistance is this special case. You cannot have fruitful discussions of series and parallel circuits without knowing what nodes, branches, etc are.

Resistances can be combined in parallel or series, this conversion can lead to an equivalent circuit or resistance which is useful for understanding. The current entering the node with one voltage source is the sum of the currents in the branches, and the currents in the individual branches are determined by ohms law, one can easily prove addition of resistances in parallel using KCL, however without any circuit to work with, its quite difficult to know what you mean exactly.

 

[CWatters]

It can’t be the current along the entire path between the two points because the current splits at junctions.

The current "I" is the total or net current flowing between the "beginning" and "end" nodes. However there does not need not be a place anywhere that a current of that value can be measured.

Consider the circuit below. It has two resistors in parallel (1R and 3R) between two nodes called "beginning" and "end". The equivalent resistance is 0.75R. The voltage between these two nodes is 1.5V so the total current according to Ohms law is I = V/R = 1.5/0.75 = 2A. The individual currents flowing in the two resistors are 1.5A and 0.5A also making 2A in total.

There is no single wire anywhere in the circuit where you can actually measure 2A.

 

[Dale]

It can’t be the current along the entire path between the two points because the current splits at junctions.

A resistor is a two-terminal device. The current in one terminal is equal to the current out the other terminal. There is no splitting. Then I is the current entering the resistor and V is the voltage of the "in" side measured relative to the voltage at the "out" side, and V=IR.

 

[ZapperZ]

So if R is the resistance between a point at the beginning and at the end, and V is a voltage between a point at the beginning and at the end, V/R would be the current coming from the beginning and entering the end of the circuit?

It can’t be the current along the entire path between the two points because the current splits at junctions.

I strongly suggest you look at the derivation of equivalent resistance in series and in parallel. This is because in such derivations, Ohm's law is applied all over the place, including what the relevant current is used in each component.

Zz.

 

[sophiecentaur]

The current is not the same throughout the whole circuit. Where is the current equal to this value?

Many of these replies to the OP are just overkill. Isn't this just the old chestnut about current being 'used up' on the way round a circuit? As far as I can see, it is and it needs to be answered at an appropriate level. The current is the same coming out of the circuit as going in. What is actually in the black box 'between the two terminals' is not relevant; current does not get used up.

 

[ZapperZ]

Many of these replies to the OP are just overkill. Isn't this just the old chestnut about current being 'used up' on the way round a circuit? As far as I can see, it is and it needs to be answered at an appropriate level. The current is the same coming out of the circuit as going in. What is actually in the black box 'between the two terminals' is not relevant; current does not get used up.

Unfortunately, this is not what the OP understood. He/she seems to think that there are "many currents", and somehow not realizing that it is only the single current going in and out of the "blackbox" is the only thing relevant TO the blackbox.

It is why I suggested that the OP look at the derivation for parallel and series circuit. In such derivations, it is clear what the "constant" values are, and how currents behave both in each resistor elements and for the entire complex.

Zz.

 

[FactChecker]

In the simple parallel circuit of the OP, each individual parallel path has a current which agrees with the resistance and the voltage drop along that path. Furthermore, the total current at an end junction agrees with the equivalent resistance (R in the equation $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$) and the voltage drop between the end junctions.

For more complicated networks, Kirchhoff's laws must be applied to understand the currents and voltages.

 

[sophiecentaur]

It can’t be the current along the entire path

It doesn't have to be. It's the current flowing between the terminals. What goes on in between (the paths) in a purely arbitrary.