There is a series curcuit and there are 5 resistors in it which are the light bulbs. I know that current is directly proportional to the voltage and inversely proportional to the resistance. Can anyone explain to me how resistance affect the voltage drop in each of the resistor in the series circuit? Thanks in advance.


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Before talking about resistance, we need to get to grips with what actually is being resisted by a component in the first place.

Which is, the flow of charge. Charge is measured in C and is an indicator for the quantity of electrons flowing through a circuit. The unit C itself stands for coulomb, which is a large group of electrons, approximately 6.24x10^18 of them.

Every component, a lamp in this case has resistance, which is a measure of the difficulty of each coulomb’s passage. This difficulty of passage is physically expressed as collisions between a component’s atoms and flowing electrons. When they collide, the electrons transfer energy to the components atoms, causing them to more, as they now have more KE to vibrate with. This increases the lamps temperature too. So when the resistance of a component is greater, each coulomb of electrons has to transfer more energy to the component’s atoms in order to get through.

This change in energy between a coulomb of electrons before passing a component and after passing a component is also known as the voltage across a component too.

Sometimes equations don’t tell a story like this, that’s because they’re often complex derivations.


Can anyone explain to me how resistance affect the voltage drop in each of the resistor in the series circuit?

You should have learned Ohm's law, which tells you the voltage drop across a (ideal, linear) resistor in terms of the current through it:


So to find out the voltage drop across a resistor, you need to know the current through it.

For the case you described, 5 resistors in series powered by (I'll assume) a constant voltage source, you will need to use two facts to solve it:

  • The current through components connected in series is equal. (This is from Kirchhoff's Current Law or KCL)

  • The total voltage drop across the 5 resistors is the same as the voltage drop across the battery (This is from Kirchhoff's Voltage Law or KVL)

For real light bulbs you'd have to deal with the fact that they are nonlinear resistors so their resistance changes pretty dramatically as the current through them changes, but you aren't likely expected to include this in the analysis if you're still at the stage of learning where you just learned Ohm's law.

  • $\begingroup$ Thanks for the great explanation; but can you explain further on how does the resistance affect the voltage drop? $\endgroup$ – Bido262 May 8 at 14:39
  • 1
    $\begingroup$ @Bido262, there's nothing more to it. The voltage drop is equal to the current times the resistance. How much more explanation do you want? $\endgroup$ – The Photon May 8 at 14:50
  • $\begingroup$ If you want the physical mechanism behind resistance, you can Google "Drude model" but realize this model ignores quantum mechanics, so is kind of out of date. $\endgroup$ – The Photon May 8 at 14:55

We know two things about the circuit. (1) the total voltage is constant and (2) the current through each of the components is the same. You didn't actually specify the voltage source so (1) might not be true but if we assume the voltage source is a battery this will be the case. You can see why (2) is true: if one component has a higher current the electrons in the wire will pile up. For the same reason cars can't travel with different speeds on a single road. If at one portion of the road the cars are faster the cars will either pile up or there will be a gap. Electrons react to this very quickly so in a series circuit the current will always be the same.

Using Ohm's law you can derive all the voltages and currents. I used three resistors to make it easier to compute. In a series circuit you can add resistors to get an effective resistance. We can use this and Ohm's law to determine the current through the circuit: $$I=\frac{V_{total}}{R_{total}}$$ Usually $V_{total}$ is known, since it just the voltage of the supply. $V_{total}$ is also $V_1-V_4$. Using Ohm's law again you can calculate the voltage drop over $R_1$: $$V_2=V_1-IR_1$$ where $IR_1$ is the voltage drop over $R_1$. You can iterate this until you reach $V_4$. If every resistance is the same every voltage drop is also the same and the voltage is equally divided over the resistors. Note that you can add a constant to all the voltages without changing the setup. You can only measure voltage drops, not absolute voltages.



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