Lets say we make two circuits, one a simple circuit connected to a 9 V battery with say a 9 $\Omega$ resistor attached,and one another simple circuit connected to a 9 V battery with a 9 $\Omega$ resistor and a light in series. Now in both the circuits the total voltage drop will be 9 V. However, in the first circuit, there will be a 9 V drop over the 9 $\Omega$ resistor, and in the second circuit there will only be a less than 9 V drop over the same 9 $\Omega$ resistor (but still a 9 V drop between each battery terminal).

My question is, how can this same 9 $\Omega$ resistor produce two different voltage drops? I would imagine that since a simple resistor has a constant resistance, it would produce a constant voltage drop. Similarly, how can the total higher resistance of the series circuit produce the same voltage drop as a lower resistance circuit, such as the 9 $\Omega$ resistor wired in series with the battery? This is all assuming a constant voltage source such as a battery.

Maybe this is just a case of me overthinking and missing something simple. As a note I am just getting into electronics.

  • 1
    $\begingroup$ It would only produce the same voltage drop if it has the same current. Are you sure that's the case here? $\endgroup$
    – JMac
    Commented Mar 6, 2018 at 16:08
  • $\begingroup$ According to your intuition you don't even need a battery. Your resistor will "Produce" the same voltage, no matter what, right? $\endgroup$
    – nasu
    Commented Jan 13, 2021 at 17:27

2 Answers 2


The light in series also has a resistance. Let's call it RL. The net resistance is RL + 9Ω. The result is that the voltage across the 9Ω resistor is

9 volts x 9Ω/(RL+9Ω), and the voltage across the light is 9 volts x RL/(RL+9Ω).


You are missing the current in your assumptions.

The resistance is calculated as $R=\frac{V}{I}$ ($V$ voltage, $I$ current). A $9\Omega$ resistor is a thing that keeps this fraction constant (at $9 \frac{\text{V}}{\text{A}} = 9 \Omega$) if it is not overloaded.

If you connect this thing to an ideal voltage source (your typical 9V block is far from that for this load), a current of 1A will flow ($I = \frac{V}{R}$). If you add another part to the circuit in series, the total resistance will rise and consequently the total current will drop.

Because $V_{\text{drop at resistor}}=R_{\text{ of resistor}} \cdot I$, this also leads to a lower voltage drop.

I think it helps to think of a voltage drop not as something that is produced but something that just happens due to a resistance.

  • $\begingroup$ I get mathematically that in the higher resistance circuit, the current decreases such that the quantity IR, the voltage, would be the same as in the other circuit. As you said, voltage drops are due to resistance. Therefore, it seems to me that a higher resistance force would cause a higher voltage drop, but if were using a constant voltage source, this can't be the case. But why not? That's what I don't get. Why would the voltage not increase? The expression V=IR seems more like a model, and doesn't tell me fundamentally why a decrease in current in a high R circuit maintains the same V. $\endgroup$ Commented Mar 7, 2018 at 2:07
  • $\begingroup$ The voltage doesn't increase because it's the constant voltage source's job to keep the voltage between its Terminals constant. An ideal voltage source does exactly that. The overall resistance then determines the current. $\endgroup$
    – Jasper
    Commented Mar 7, 2018 at 8:08
  • $\begingroup$ . You use the phrase "resistance force". Can you explain what you mean? If anything, the force that a resistor applies to the charge carriers is a friction-like force that opposes the motion of the charge carriers. $\endgroup$
    – garyp
    Commented Apr 24, 2020 at 21:02

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