I have looked this up and what followed was that resistance does not affect the voltage of a circuit. But lets say we have a power source of 5V flowing at 1 Amp. The neccesary resistance of this circuit must be 5 $\Omega$ to decipate the energy of the eletrons before they get to the negative terminal of the source. So this is referred to as the voltage drop across the resistor but has this not affected the voltage by reducing the energy of the flowing charge?
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$\begingroup$ Is your power source strictly 5V at the output with only 1 A of current, all the time? Where do you get that power source, because it's extremely unusual. $\endgroup$– Bill NCommented Dec 28, 2021 at 14:45
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3 Answers
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"[R]esistance does not affect the voltage of a circuit."
I suspect that the writer had in mind what happens when you connect different resistances across a power supply. A good power supply is built to give a roughly constant voltage across its terminals – a voltage that hardly changes when you change the resistance of the load (that is whatever it is that you connect across the terminals) – provided that you don't make the load resistance too low.
A lower load resistance, $R$ will just increase the current according to $$I= \frac {V_{\text{terminal}}}{R_{\text{load}}}$$
Even an ordinary battery will provide a roughly constant voltage across its terminals, provided that you don't connect a load that has too low a resistance across its terminals. In fact the terminal voltage is given, fairly accurately, by $$V_{\text{terminal}}=\mathscr E - Ir$$ $\mathscr E$ is called the emf of the battery, and is the voltage across its terminals when you are drawing no current. $r$ is called the internal resistance of the battery.
Example: Suppose a battery has an emf of 9.0 V and an internal resistance of 0.5 ohm. What will be the terminal voltage if you connect a 4.0 ohm resistor across its terminals? $$V_{\text{terminal}}=9.0\ \text V - I \times 0.5\ \Omega \ \ \ \ \ \ \text{but}\ \ \ \ \ \ \ V_{\text{terminal}}=I \times 4.0\ \Omega$$ Solving, $I=2.0$ A, $V_{\text{terminal}}=8.0$ V. If you repeat for an 8 ohm resistor connected across the terminals of the same battery, you will find that $V_{\text{terminal}}$ isn't much different!
answered Dec 28, 2021 at 12:54
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Yes, resistance does affect voltage. In fact, there is an equation that links voltage, current and resistance - $V=IR$, where $V$ is the voltage in volts, $I$ is the current in amperes and $R$ is the current in ohms ($Ω$). Whoever wrote that resistance does not affect voltage has made a terrible mistake.
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$\begingroup$ Interesting....But doesn't this imply that as resistance increases so does voltage for constant current? That seems very counterintuitive $\endgroup$ Commented Dec 28, 2021 at 11:59
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$\begingroup$ @SwissGnome Why does it seem counterintuitive? It would be strange if current increases when resistance also increases. $\endgroup$– MathGeekCommented Dec 28, 2021 at 12:05
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$\begingroup$ True. Is it even possible to maintain constant current when increasing resistance since current is inversely proportional to resistance? $\endgroup$ Commented Dec 28, 2021 at 12:09
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$\begingroup$ @MathGeek If the source of voltage is reliable, the voltage does not drop with increased current. If one decreases the resistance $R$ the current just increases as $I=V/R$. Unless the resistance is so small that is amounts to a short-circuit. The current becomes so huge that eventually the voltage does drop... and you start an electric fire... $\endgroup$– AlfredCommented Dec 28, 2021 at 18:01
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You were probably reading about a circuit connected to an “ideal” voltage source, I.e. a voltage source having no source impedance that would be in series with the connected circuit resistance.
But all real voltage sources have impedance causing a drop in the source voltage when current flows. Lowering the external resistance draws more current causing a further decrease in the source voltage.
However, if the source impedance is much less than the circuit resistance, the voltage drop would be negligible.
Hope this helps.
