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schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

Are these two circuits the same? If not, could you please indicate why as well?

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  • \$\begingroup\$ What do you think, and why? \$\endgroup\$ – John D May 29 at 21:33
  • \$\begingroup\$ How many path for a current to flow do you see in the first circuit? \$\endgroup\$ – G36 May 29 at 21:35
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When starting out, your brain is easily tricked by the arrangement in which things are drawn rather than what they actually are. Side-by-side doesn't automatically mean parallel.

What if I took your "parallel" circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

and just changed it to this. Does it still look like it is in parallel to you? Or series for that matter? Remember, the current can flow into the circuit on those stubs from outside.

schematic

simulate this circuit

  • Parallel = voltage across all components are the same
  • Series = current through all components is the same

Don't be tricked by how things are arranged on a page.

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Whatever way you draw it it's a series circuit and the total resistance is R1 + R2.

schematic

simulate this circuit

Figure 2. R1 and R2 are in parallel in this case.

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  • \$\begingroup\$ *As long as those wire stubs aren't there. \$\endgroup\$ – DKNguyen May 29 at 22:04
  • \$\begingroup\$ The OP question was about whether or not the two circuits were the same, and @Transistor responded in a very fine way, showing also clear pictures. \$\endgroup\$ – barrow Jun 3 at 9:53
  • \$\begingroup\$ The most important point here is that with Thevenin, often people hasn't clear the concept of "circuit equivalence" and, mainly, that the "equivalence" need to be applied between 2 points/nodes. \$\endgroup\$ – barrow Jun 3 at 10:00
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There is a good way to understand either a combination is series or parallel. For example, take a look at the circuit below.

schematic

simulate this circuit – Schematic created using CircuitLab

The left side of R1 is directly connected to the positive terminal of the battery. Let's say the voltage is Va. Same as this, the right side of R1 is connected to the negative terminal of the battery. Let's say the voltage is Vb.

So the potential difference across R1 is (Va - Vb)
Similarly voltage across R2 will also be (Va - Vb)

If two resistors are parallel, then they will have the same potential difference between their terminals.

schematic

simulate this circuit

Now for this new circuit, the upper side of the resistor R1 is directly connected to the positive terminal of the battery. The voltage of that terminal is labeled as Va.

Now there will be a voltage drop caused by the resistor R1 (N.B. Ohm's law V=IR). So a new voltage level Vc will appear at the lower side of R1. And the lower side of R2 is directly connected to the negative terminal of the battery, which is labeled as Vb.

All series combination will cause this type of voltage change.

Let's take a complicated example.

schematic

simulate this circuit

Just by inspection, it's quite difficult to understand. But if the node voltages are marked, then you will see that all of them have the same potential difference, which is (Va - Vb). This means that they are all parallel to each other.

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Sometimes, when you're still new to all this, it just isn't clear if two circuit elements are parallel (or series) connected.

If you're not sure, there's an easy 'trick' to test whether or not the two resistors are parallel connected: consider setting one of the resistors to zero ohms.

For example, set \$R_1=0\Omega\$ (essentially, replace \$R_1\$ with a wire).

It's clear that the voltage source sees the resistor \$R_2\$ as the load.

But this wouldn't be the case if \$R_1\$ and \$R_2\$ are parallel connected since, with \$R_1=0\$, we have

$$R_1||R_2 = \frac{0\cdot R_2}{0 + R_2} = 0\Omega$$

That is, zero ohms in parallel with \$R\$ ohms equals zero ohms.

Since the source doesn't see zero ohms with \$R_1=0\Omega\$, it follows that the two resistors are not parallel connected.

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Maybe your confusion comes from the Thevenin equivalent circuit of the one you show (the voltage source Vth=V*R2/(R1+R2) in series with the resistor Rth=R1//R2). but you don't have to short circuit the output.

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These are two different circuits because in first circuit the current is same throughout the circuit but in 2nd circuit if we place R1and R2 in parallel than current is divide between the resistors by the rule of KCL.

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