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I'm struggling with the theory of this(unusual) circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The load receives no power at all, due to the voltages cancelling out, yet the parallel sub-circuits voltages do not cancel out, and enormous current flows in each sub circuit. I dont understand why current in the whole circuit(even in the sub-circuits) is not equal to zero? If the voltage at node Vx is V1 - V4 = 0, likewise with Vy being V4 - V1 = 0.

I'm trying to understand how this is physically possible(for current to flow or shorting out in the sub-circuits) while the potentials are opposing one another, I tried looking at each branch and how it's possible for current to flow there but it's just difficult.

The reason why I think current flows in the sub-circuits, is because of them being shorted. But still lost with the concept while analyzing.

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  • 1
    \$\begingroup\$ The answer is simple. If there is a difference in potential (in voltage) between two points and if there is a path for a current to flow the current will flow. But if potential difference is 0 no current can flow. In your sub-circuits 1 there is a potential difference between V1 and V2 (6V) so the current must flow 6V/1 miliohm = 6000A \$\endgroup\$ – G36 Jun 21 '16 at 16:16
  • \$\begingroup\$ It flows in a loop around each sub-circuit. \$\endgroup\$ – immibis Jun 22 '16 at 4:34
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Step 1) Use Thévenin's Theorem to mathematically derive the Thévenin equivalent circuit for subcircuit V1, V2, Ri_1, and Ri_2. (Hint: The Thévenin equivalent should be 6 VDC in series with 250 μΩ).

Step 2) Use Thévenin's Theorem to mathematically derive the Thévenin equivalent circuit for subcircuit V3, V4, Ri_3, and Ri_4. (Hint: The Thévenin equivalent should be 6 VDC in series with 250 μΩ).

Step 3) Redraw the circuit, replacing the two subcircuits with their Thévenin equivalent circuits. From the Load resistor's perspective, this simplified circuit is equivalent to the original circuit.

Step 4) Use Kirchhoff's Voltage Law (KVL) to calculate the current in the simplified and equivalent series circuit. (Hint: Your calculated current should be zero amps.)

So the Load resistor has zero amps flowing through it, and each subcircuit (in the original circuit) is a 6 kA current loop.

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TL;DR: I wasn't going to bother doing a full derivation and settle with a handwavy argument that we know the load current is zero because the two sub-circuits are identical and so the voltage across the load resistor will be zero, but I decided that it would probably not help very much, so full derivation it is.


Lets ignore the ground node that you have placed - there is only one of it, so it is doing nothing. Also, lets label the node between \$R_1\$ and \$R_2\$ as \$V_u\$, and the node between \$R_3\$ and \$R_4\$ as \$V_w\$.

It is then worth a reminder of what a potential difference is. It is the difference in voltage between any two nodes. The 9V battery has a potential difference at its positive terminal of 9V referenced to its negative terminal.


Lets first consider the sub-circuit individually - you can't always do this, in fact frequently you can't, but because the two are identical, I know that I can. You have 6V dropped across effectively a 1mOhm resistor. As a result you are going to get large currents - 6kA worth.

Lets try to work out the voltage at the node \$V_u\$ that I defined earlier with respect to node \$V_x\$, or to put it another way, the potential difference \$V_{ux}\$.

Well, we know from Kirchoffs Current Law that the sum of all currents on all braches at a node equals zero. So what are the currents into the node U? Well, there is the load current, the current through \$R_1\$ and the current through \$R_2\$. Lets put this in a calculation:

$$I_{R_1} + I_{R_2} + I_{load} = 0$$

So what are these currents. Well, lets work them out. First \$I_{R_1}\$

$$I_{R_1} = \frac{V_{R_1}}{R_1} = \frac{V_u - (V_x-9)}{R_1} = \frac{V_u - V_x + 9}{500\mu\Omega} $$

Then \$I_{R_2}\$

$$I_{R_2} = \frac{V_{R_2}}{R_2} = \frac{V_u - (V_x-3)}{R_2} = \frac{V_u - V_x + 3}{500\mu\Omega} $$

Bringing that all together:

$$I_{load(u)} = -\frac{V_u - V_x + 3}{500\mu\Omega} - \frac{V_u - V_x + 9}{500\mu\Omega} = \frac{-(V_u - V_x + 3) - (V_u - V_x + 9)}{500\mu\Omega} = \frac{-2V_u + 2V_x - 12)}{500\mu\Omega} = \frac{V_x - V_u -6}{1m\Omega}$$

Do the same with the other subsystem - you will get the same equation but with \$V_y\$ and \$V_w\$ instead - i.e:

$$I_{load(v)} = \frac{V_y - V_w -6}{1m\Omega}$$


Now we have analysed each subsystem, lets calculate what the load current is. We know that the current flowing through load resistor will be equal in magnitude on each side - because currents through a series branch are always equal.

$$\begin{align} I_{load(u)} &= I_{load(v)}\\\\ \frac{V_x - V_u -6}{1m\Omega} &= \frac{V_y - V_w -6}{1m\Omega}\\\\ V_x - V_u &= V_y - V_w \tag{1}\\ \end{align}$$

If we look at the connection between \$V_x\$ and \$V_y\$ we can see that they are directly connected. That is to say \$V_x = V_y\$. So, substituting that in to (1), we get:

$$\begin{align} V_y - V_u &= V_y - V_w\\\\ V_u &= V_w \tag{2}\\ \end{align}$$

We know then work out the current through the load is:

$$I_{load} = \frac{V_{load}}{R_{load}} = \frac{V_{uw}}{2k\Omega}= \frac{V_{u}-V_{w}}{2k\Omega} \tag{3}$$

So substituting (2) into (3) we get:

$$I_{load}= \frac{0}{2k\Omega}=0A$$

We can see then that there is no current through the load.


You can also then work out the current flowing in each sub-circuit. Remember we now know that the load current is zero which means each sub-circuit is effectively independent. You should be able to work out that the voltage over \$R_1\$ and \$R_2\$ is \$9-3 = 6V\$. So we can say that the current is:

$$I = \frac{6}{R_1+R_2} = \frac{6}{1m\Omega}=6kA $$

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I dont understand why current in the whole circuit(even in the sub-circuits) is not equal to zero?

Step by step:

(1) Consider just one of the sub-circuits:

schematic

simulate this circuit – Schematic created using CircuitLab

Clearly, there is a current \$I_1\$ circulating clockwise equal to

$$I_1 = \frac{9\mathrm{V} - 3\mathrm{V}}{0.5\mathrm{m\Omega} + 0.5\mathrm{m\Omega}} = \frac{6\mathrm{V}}{1\mathrm{m\Omega}} = 6\mathrm{kA}$$

(2) Consider two of these sub-circuits that are not connected. They both have \$6\mathrm{kA}\$ current circulating clockwise.

(3) Connect a wire between these two sub-circuits as follows:

schematic

simulate this circuit

This does not change the operation of the independent sub-circuits, however, we can now define the voltage between the two bottom nodes.

Now, if there is a non-zero voltage difference between the two bottom nodes, one node is more positive than the other.

Thus, if we exchange the position of the two sub-circuits, the polarity of the voltage across the two bottom nodes will reverse.

But consider; since the two sub-circuits are identical, it is impossible to distinguish between the original configuration and the swapped configuration and thus we should not expect the voltage between the two bottom nodes to change if the sub-circuits are swapped.

The only voltage consistent with this is \$0 \mathrm{V}\$ and so we conclude that there is zero volts between the bottom two nodes.

A load resistor placed across zero volts will have zero current through and so, connecting a load resistor between the bottom two nodes does not change the operation of the circuit.

In summary, both sub-circuits have \$6\mathrm{kA}\$ current circulating clockwise but there is zero current through the load resistor for the reason that the two sub-circuits are identical.

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Easiest way to understand is to consider effect of each supply individually.

Using superposition.

schematic

simulate this circuit – Schematic created using CircuitLab

\$500\mu\ \parallel\ 500\mu\ = 250\mu\$

\$250\mu \ s\ 2k \approx 2k\$

\$2k \parallel\ 500\mu\ \approx 500\mu\$

\$500\mu \ s\ 500\mu = 1m\$

\$I_T = \frac{V_1}{R_T} = \frac{9V}{1m\Omega} = 9kA \$

Current flow is inversely proportional to resistance, so the majority flows through \$Ri_2\$ with approximately: \$ I_{Load_1\ (MAX)} = \frac {500\mu\Omega}{2k\Omega}\ \times \ 9kA = 2.25mA\$ (current division) going through the load resistor.

Now 2.25mA on 9kA is meaningless (0.000025%). But 2.25mA on the load resistor would be 4.5V.

Similarly: The effect of V2 is \$I_T = 3kA \$, \$ I_{Load_2} = 0.75mA\$. Equivalent resistance is the same.

V1 and V2 work against each other so current in each branch is 9kA - 3kA = 6kA (ignoring 3mA going through load).

Same analysis works for V3 and V4, but polarity changes, so:

$$I_{Load} = 2.25mA + 0.75mA - 2.25mA - 0.75mA = 0A$$

Or:

$$V_{Load} = 4.5V + 1.5V - 4.5V - 1.5V = 0V$$

But each source has 6kA flowing through them. All the power is wasted in sources.

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  • \$\begingroup\$ "Current flows through the path of least resistance" - this isn't true. \$\endgroup\$ – Alfred Centauri Jun 22 '16 at 1:53

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