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In the following diagram voltage can exist without current since it's an open circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Can a voltage source in an open circuit with 0 current effect a circuit like so:

schematic

simulate this circuit

So that current would be equal to what is outputted from V-1, yet the voltage is increased to 15V due to V-2?

Because in this circuit voltage can increase even if it's an open circuit:

schematic

simulate this circuit

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  • \$\begingroup\$ The negative terminal of V2 is basically your reference point for measuring 15 volts. Since that reference point is not connected to anything else in the circuit, it won't affect its operation (ideal case). \$\endgroup\$
    – hryghr
    Nov 15 '14 at 17:16
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The clue to voltage is in it's proper name: Potential Difference.

Voltage isn't a real thing, it's just the difference between two things.

Imagine, if you will, a tank of water. Say it's 1m on each side, and you fill it to a depth of 10cm. You lift the tank up off the ground to a height of 1m. So the top of the water is 1.1m above the ground.

Now you make a hole in the bottom of the tank. The water flows out of the hole.

Until you had made that hole (created a circuit) the difference in heights still existed.

Now imagine the same scenario, but you instead fill the tank right to the top. The top of the water is now 2m above the ground. You have a greater potential difference between the top of the water and the ground.

Make the same hole, and the water comes out with more force due to the increased weight of the water in the tank - the greater potential difference causing greater current to flow.

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  • \$\begingroup\$ Thank you for that great illustration. But my concern is in the second circuit is 15V true? Since voltage could exist even in an open circuit? \$\endgroup\$
    – Pupil
    Nov 15 '14 at 17:39
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    \$\begingroup\$ @key No - in the 2nd circuit the 5V supply is totally irrelevant to any current flow thru the load from the 10V source. If I stood on suitable insulating blocks with a multimeter in my hand across a 9V battery and then someone wired me to 1-million volts (i.e. with a van der graaf generator), the meter would still read 9 volts. \$\endgroup\$
    – Andy aka
    Nov 15 '14 at 17:52
  • \$\begingroup\$ Ah thanks, I assumed that I could "add" voltage without current. Yet it seems pointless thank you! \$\endgroup\$
    – Pupil
    Nov 15 '14 at 18:01
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Your second diagram is deceiving because pointing to a single point, and saying "N volts" is basically meaningless by itself.

For voltage to make any sense, you need to specify two points. A voltmeter (for example) has two inputs, and you connect one of those inputs to each of those two points.

Now, it's true that in many circuit diagrams you'll see things like +5V or 3V3, or things like that. When you see this, however, it means there's some implied reference point, usually marked as the ground. So, a marking like "5V" really means "this point will be at 5 volts above ground".

Here, the point you've marked as "V=15" depends on what you choose as the reference point (the ground, if you will). For the moment, let's leave the resistor out of the circuit, since it's (mostly) irrelevant to the question at hand. Instead, let's consider something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

If we measure between A and B, we'll see 10 V. If we measure between B and C, we'll see 5 volts. If we measure between A and C, we'll see 15 volts. The voltage between B and C makes no difference to the voltage between A and B (and vice versa).

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