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I have this resistor which is connected to some battery (not very important). I don't understand why the total voltage is equal to the voltage drop over the 3 \$ \Omega \$ and 2 \$ \Omega \$ resistors. Why is it not dependent on the resistor in the middle with 4 \$ \Omega \$ . I do understand it if the 4 \$ \Omega \$ were substituted with a wire. But how can it be that there is no difference whether it is 4 \$ \Omega \$ or 0 \$ \Omega \$?

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    \$\begingroup\$ "(not very important)": oh, yes, it is. Because how you attach your battery changes the way you look at this circuit. Because if you attach one contact of the battery to the left-hand stub and the other contact to the right-hand stub, none of the resistors matter: the battery defines the voltage. The resistors just define the current that flows in result. So, draw your battery into this and think hard about what a connection ---- in your drawing means. \$\endgroup\$ – Marcus Müller Jan 18 '20 at 18:22
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I don't understand why the total voltage is equal to the voltage drop over the 3 Ω and 2 Ω resistors.

Normally on these exercises we are dealing with ideal voltage sources. This means that they will hold their output voltage no matter what the load is. That means that any pattern or combination of resistors can be connected across the voltage source and it will stay constant. (Any value except 0 Ω directly across the battery terminals since that will give an invalid result for \$ I = \frac {V}{R} \$.)

What will change is the current. Removing, changing or short-circuiting the 4 Ω resistor will affect the current drawn from the voltage source.

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  • \$\begingroup\$ Does this mean that the voltage over the 1 Ω 4 Ω and 2 Ω resistors will be the same? \$\endgroup\$ – dfphysicsi Jan 18 '20 at 18:46
  • \$\begingroup\$ @dfphysicsi, pick any two nodes you like in the circuit, and label them A and B. Whatever path you go by from A to B, the total drop will be the same. This is just another way of stating Kirchhoff's Voltage Law (KVL). \$\endgroup\$ – The Photon Jan 18 '20 at 19:08
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If you are familiar with Kirchoff's Voltage Law, then you know that any path that connects your two (ideal) battery nodes (assumed to be at the far left and the far right of your diagram) has the same total voltage across it as the battery. Thus the 2R and the 3R path has the same voltage as the battery. The same is also true for the path containing the 5R and the 1R. The 4R does not enter into this at all, which could be replaced by any other value without changing the result.

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  • \$\begingroup\$ I am sorry, I don't quite follow your conclusion. I just don't uderstand why the 2R and the 3R path has the same voltage as the battery. How is it that the 4R does not enter into this? \$\endgroup\$ – dfphysicsi Jan 18 '20 at 18:31
  • \$\begingroup\$ Draw a closed loop that contains the battery, 3R and 2R, and then apply KVL to the closed loop. \$\endgroup\$ – Phil Freedenberg Jan 18 '20 at 18:36

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