How can same points have different current?

Question

Aren't B and D the same point and shouldn't it actually be (B and B)?

If they aren't the same point what should be the current from B to D?

Why is it 2 amperes (from the answers) and not 10 amperes as it should be (because the two 1ohm and 4 ohm connected in parallel are in series with the 2 ohm and 3 ohm in parallel?

Another question is why should there be any current through the 3ohm resistance at all? Isn't the wire a short circuit and shouldn't current flow through the short circuit across the the 2ohm as it offers less resistance? What is actually going on?

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If I can calculate the total current using the 2nd circuit as equivalent to the 1st, doesn't that mean that 10 A is flowing through BD? Can I take the red wire as BD or can I not? If not, why? Where is BD in the 2nd (equivalent) circuit? Isn't BD the same point in the equivalent circuit? Or do I have to come back to the 1st circuit after calculating the current from the 2nd (equivalent) circuit? If yes, why do I have to come back to the 1st circuit? Can I not have my BD wire in the 2nd circuit? If not, how is it the equivalent circuit of the 1st one in the first place?

Answer 1

Aren't B and D the same point and shouldn't it actually be (B and B)?

If you were analysing voltages then yes, B and D are the same node on the circuit. In this case, however, we're looking at current flow so different labels are applied to make it clear which points in the circuit we are referring to.

If they aren't the same point what should be the current from B to D?

That's what you are being asked to calculate.

Why is it 2 amperes (from the answers) and not 10 amperes as it should be (because the two 1ohm and 4 ohm connected in parallel are in series with the 2 ohm and 3 ohm in parallel?

• 4 Ω || 1 Ω = 0.8 Ω.
• 2 Ω || 3 Ω = 1.2 Ω.
• 0.8 Ω + 1.2 &Omega = 2 Ω (the resistance seen by the 20 V source).
• The current from the 20 V source = 20 / 2 = 10 A.

The 10 A current is split between the parallel resistors and since the ratios of the upper pair and lower pair are not equal some current will flow through the BD bridge.

Another question is why should there be any current through the 3ohm resistance at all ? Isn't the wire a short circuit and shouldn't current flow through the short circuit across the the 2 ohm as it offers less resistance? What is actually going on?

You seem to be falling into the "current takes the path of least resistance" error. It doesn't. Current in each branch is inversely proportional to its resistance. That is, current will flow in every branch but more will flow in the lower resistance branches.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. The simulation result for the circuit with ammeters in each branch.

Note that the BD branch has 8 + 2 = 10 A coming in from the left and 6 + 4 = 10 A exiting on the right. The BD link passes the current difference from one side to the other.

^{simulate this circuit}

Figure 2. If the BD link is replaced by a bar you will still have a 2 A current drifting across the bar.

Answer 2

First calculate R1 || R2, the total resistance formed by the parallel resistors R1 and R2.

$$R1 || R2 = \frac{R1 \times R2}{R1+R2} = \frac{1 \times 4}{1 + 4} = \frac{4}{5}\Omega$$

Then R3 || R4

$$R3 || R4 = \frac{R3 \times R4}{R3+R4} = \frac{2 \times 3}{2+ 3} = \frac{6}{5}\Omega$$

The total resistance is then

$$\frac{4}{5}\Omega + \frac{6}{5}\Omega = 2\Omega$$

The total current is

$$\frac{20V}{2\Omega} = 10A$$

The voltage across R1 or R2 is

$$V_{R1} = \frac{\frac{4}{5}}{2}\times 20V = 8V$$

So, the current through R1 is 8A, the current through R2 is 2A.

The voltage across R3 or R4 is 12V.

So the current through R3 is 6A and the current through R4 is 4A.

Kirchhoff's Current Law (KCL) says the algebraic sum of all the currents flowing into a point is 0. The currents flowing into point B are the current through R1, the negative of the current flowing through R3, and the negative of the current flowing from B to D. So the current flowing from B to D is 8A - 6A = 2A.

Applying KCL to point D, gives the current from B to D as -(2A - 4A) = 2A. As one would expect, the answer is the same whether you choose to apply KCL to point B or to point D.

Answer 3

First, search the potential on center point B -> Vbc =12 V, and Vab=8V.

Current in 1 Ohm is 8 A. Current in 4 Ohm = 2 A.
Current in 2 Ohm is 6 A. Current in 3 Ohm = 4 A.

Do the balance and you find the current through B-> D = 2A.

It is not the same result if 1 and 4 Ohm are tied together in series with 2 and 3 Ohm tied together.

Answer 4

You've combined the resistances correctly to find the 10A flowing through the combination. Now, use the current divider equation to split the currents flowing through each parallel combination of resistances. Note that the phrase "path of least resistance" is misleading since the figurative interpretation implies that only one path is "traveled" while the literal, electrical interpretation is somewhat different.

Answer 5

Aren't B and D the same point and shouldn't it actually be (B and B)?

From a voltage point of view, you are right.

I deal with the node naming issue the following way. Voltage nodes are labeled with a lowercase letter. Current nodes always have an associated voltage. A voltage node may have one or more current nodes. So I name current nodes as a numerical subscript to a voltage node. This allows all current nodes within the same voltage node to have the same letter designation indicating the voltage while having a unique label. So your nodes B and D become b1 and b2 respectively.

For branches I use the label for current as the label from the branch since the definition of a branch is a cascade of elements in series and so all have the same current called the branch current. The other answers have calculated the currents, so I won't do it again. What I want to discuss is the equivalences and differences between the two configurations, shown below. So I hope the labeling I have chosen will help clarify.

The two configurations are not the same but they are equivalent for the for the following reasons.

1. The load seen by the voltage source is the same.
2. The voltages and currents across and through all the resistors are the same.
3. The node voltages measured to an arbitrary reference are all the same.

Making these measurements cannot identify the configuration. Example: The labeling used allows \$V_{bc}\$ to mean the same in both configurations.

What is different between the two configurations is the current nodes within voltage node \$b\$. For clarity the current nodes in the first configuration are labeled \$b_{1}\$ and \$b_{2}\$, while in the second configuration they are labeled \$b_{3}\$ and \$b_{4}\$. They are not the same nodes as KCL will identify.

Measuring the current flows in and out of the node within node \$b\$ will identify the configuration.

So what happened to \$b_{1}\$ and \$b_{2}\$ in the second configuration?

Nothing happened. They just don't exist there. I don't agree that there is some sort of drifting electrons across the diameter of the conductor. Testing with magnetometer should be able to answer this question.

^{simulate this circuit – Schematic created using CircuitLab}

^{simulate this circuit}

Another question is why should there be any current through the 3ohm resistance at all ? Isn't the wire a short circuit and shouldn't current flow through the short circuit across the the 2ohm as it offers less resistance? What is actually going on?

Two things you can always count on:

1. Current leaving one end of a component must enter the other end of the component.
2. Charge always flows in a closed path regardless whether it is easy or tortuous.

Really these two are different takes on conservation of charge.

It is always a good endeavor to identify the path(s) the charge flow takes to return to its source. It will change direction, divide or split at various nodes.

I hope my notation will help in understanding currents within a network of components.

Answer 6

I think your confusion stems from the fact that you correctly recognized B and D to have the same potential, meaning that for the sake of analyzing the rest of the circuit you can consider them the same node. However, that does not mean that there is no current flowing between them. In fact, to get to the zero potential difference, it is likely that some current needs to flow.

There are lots of ways to figure out that current. You can determine the total resistance (as you have done), calculate the total current from there, and then noticing, like you did, that the total current is 20V/((AB|AD)+(BC|DC)), split that total current into the currents AB and AD, and also split the same total current into the currents BC and DC. Now you'll notice that the currents AB and BC are not identical, and it is the difference that needs to be accounted for by a current running from B to D.

Another possibility is to remove the connection BD and then calculate the resulting voltage on nodes B and D, and the respective total resistance between nodes B and D (for calculating that resistance, replace every voltage source with a short and every current source with an open connection).

Then the current after connecting B and D will be this voltage difference divided by the resistance between those points, a rather interesting property of electric networks that is by no means trivial to prove but very handy to use.

Answer 7

First of all you dont have any resistors in parallel or in series.Lets solve your circuit using Y-Δ transform.

^{simulate this circuit – Schematic created using CircuitLab}

Lets focus on this part of the circuit:

^{simulate this circuit}

This is a Δ configuration circuit and it is equivalent to a Y configuration circuit:

^{simulate this circuit}

\$R_{a} = \frac{4\cdot1}{4+1+0} = \frac{4}{5}\Omega \$

\$R_{b} = \frac{0\cdot1}{4+1+0} = 0\Omega \$

\$R_{c} = \frac{0\cdot4}{4+1+0} = 0\Omega \$

so the circuit can be redrawn like this:

^{simulate this circuit}

But 0Ω means short circuit and when you replace Rb and Rc with a short circuit you find the equivalent circuit.

Questions:

Another question is why should there be any current through the 3ohm resistance at all ? Isn't the wire a short circuit and shouldn't current flow through the short circuit across the the 2ohm as it offers less resistance?

If you see the equivalent circuit I have drawn obviously the current inside the 3Ω resistor is not 0.

Why is it 2 amperes (from the answers) and not 10 amperes as it should be (because the two 1ohm and 4 ohm connected in parallel are in series with the 2 ohm and 3 ohm in parallel?

You dont have any resistors in parallel or in series.

Aren't B and D the same point and shouldn't it actually be (B and B)?

They are the same point.Take Ohm's law \$ I = \frac{V}{R} \$

If both \$ V \$ and \$ R \$ are 0 then we get a indefinite result \$ \frac{0}{0} \$ so any current could be flowing inside a short circuit.