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Resistors are used to reduce current in order to prevent light bulbs and other electrical components from “exploding”, but it is also said that “current remains the same at all points in a series circuit”.

Then what's the point of using resistor when the current will remain the same, and as it would remain the same, the light bulb would still receive that high current which could cause it to explode.

So, again, why are we using resistors? If current is still going to be the same, what's the point? The light bulb would still explode if it will not be receiving a lower current. Also, I'm talking about the series circuit, not a parallel one.

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  • $\begingroup$ The current is the same everywhere in a given series circuit. There are many types of series circuits .... each with their own current level. $\endgroup$ – PhysicsDave Aug 2 '19 at 14:17
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Note: I'm amending my answer after reading a comment from the OP that reveals (what I believe is) at the heart of the OP's question - a misconception of what is meant by "Resistors are used to reduce current...".

Right,so if I use a battery with a high voltage which would then result in high current suppose 1.0Amps this could cause harm to a light bulb due to the high current flow so I add a resistor which would then resist the flow, if the current would remain same even after it passes through a resistor wouldn't that still cause the lamp to burn out? Like before the resistor the current was 1.0 amps and I add a resistor to prevent the bulb from bursting due to the high current but after the resistor the current is still 1.0 amps then what's the point ?

Adding the resistor reduces the current everywhere in the circuit by the same amount. I might be wrong, but the bolded text leads me to believe that you're thinking of a resistor as something that 'lets less current out than is coming in'. But that's not how a resistor works.

The current leaving a resistor equals the current entering the resistor. When we say that a resistor reduces the current in a series circuit, we don't mean that it reduces the current out of the resistor compared to the current into the resistor. We mean that the current everywhere in the series circuit is reduced (by the same amount) when a resistor is added.

If, without a resistor in the circuit, the series current would be $1.0\,\mathrm{A}$, and if the resistance of the added resistor equaled the resistance of the light bulb, the current with the resistor in the circuit would be reduced to $0.5\,\mathrm{A}$ everywhere in the circuit.

It is for this reason that we can say "(series) resistors are used to reduce current".


Original answer below:

it is also said that "current remains constant in a series circuit"

What do you think that this means? While I don't care for the wording of the quoted statement, the essential idea is the current is uniform (everywhere the same value) in a series circuit$^*$

It doesn't mean that the series current doesn't change if a series resistor is added to the circuit. Adding a series resistor increases the series resistance which decreases the series current (when driven by a voltage source).

Also, I did a quick Google search on the phrase "current is constant in series circuit", and here are quotes from the top links returned:

  • The current in a series circuit is everywhere the same.
  • In a series circuit, the current is the same for all of the elements.
  • The amount of current is the same through any component in a series circuit.
  • The same current flows through each part of a series circuit

As you can see, none of these use the word constant to describe the current in a series circuit.


$^*$though not particularly important here, I should point out that it's not just that the current is the same value (since two identical parallel connected resistors have the same value of current through), it's that series connected circuit elements have identical current through.

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You seem to be proposing that all series circuits have the same current. That isn't true.

The equivalent resistance of a set of resistor in series is $$R_{eq}=\sum_iR_i$$ i.e. you just find the sum of all the resistances. If they are all ohmic then the current that flows through the circuit with voltage source $V$ is given by $$I=\frac{V}{R_{eq}}$$ Therefore, the more resistors you add the smaller the current through the circuit will be.

The power dissipated by a lightbulb in this circuit is given by $P=I^2R$, where $R$ is the lightbulb's resistance. Therefore, a lower current means less power dissipation, and less of a chance of exploding.

The current is constant throughout a series circuit (i.e. at all points in the circuit). That doesn't mean all series circuits have the same current no matter what. Certainly adding more resistance changes the current. That is why is is called resistance. It resists the flow of current.

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You misunderstood the phrase "current remains constant in a series circuit".

Consider the circuit below with a light-bulb and a resistor in series.
circuit 1
In this circuit the current through the light-bulb is the same as the current through the resistor. This is meant by "the current remains constant in a series circuit".

The current according to Ohm's law is: $$I = \frac{V_{\text{battery}}}{R_{\text{bulb}}+R_{\text{resistor}}}$$ Notice that this current is the same all the way around the circuit, because the current cannot leave the wire anywhere.

Thus the electrical current behaves very much like the water current flowing through a pipe. The main difference between both is: The electrical current is measured in Ampere, and the water current is measured in liter/second.


Now consider the next circuit with a light-bulb only.
circuit 2
In this second circuit the current through the light-bulb is greater than the current through the light-bulb in the first circuit. This is because the total resistance (light-bulb only) is lower than in the first circuit (light-bulb + resistor).

Now the current according to Ohm's law is: $$I = \frac{V_{\text{battery}}}{R_{\text{bulb}}}$$ which is greater than the current in the first circuit.

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  • $\begingroup$ Okay,so suppose I place an ammeter a few centimeters away from the negative terminal and it reads 0.8Amps but if I place it after the resistor it would be less right? Like maybe 0.5Amps ? I might be completely wrong btw! $\endgroup$ – Christina Aug 2 '19 at 13:13
  • $\begingroup$ @Christina No. The ammeter would read the same, regardless whether you put it between battery and bulb, or between bulb and resistor, or between resistor and battery. $\endgroup$ – Thomas Fritsch Aug 2 '19 at 13:19
  • $\begingroup$ Right,so if I use a battery with a high voltage which would then result in high current suppose 1.0Amps this could cause harm to a light bulb due to the high current flow so I add a resistor which would then resist the flow, if the current would remain same even after it passes through a resistor wouldn't that still cause the lamp to burn out? Like before the resistor the current was 1.0 amps and I add a resistor to prevent the bulb from bursting due to the high current but after the resistor the current is still 1.0 amps then what's the point ? $\endgroup$ – Christina Aug 2 '19 at 13:51
  • $\begingroup$ @Christina I have added some additional explanations to the answer. $\endgroup$ – Thomas Fritsch Aug 2 '19 at 14:13
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    $\begingroup$ @Christina, it's crucial to understand that, after adding the resistor, the value of the current has changed (it is now less than 1.0 amps) and the value of the current is the same everywhere in the circuit. When we say that a resistor reduces the current in a series circuit, we mean that it reduces the current everywhere in the circuit by the same amount. It's not that the current is 1.0 amp into the resistor and less out of the resistor. $\endgroup$ – Alfred Centauri Aug 2 '19 at 14:44
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Referring to the circuit drawings by @Thomas Fritsch, let's regard the voltage source as ideal, i.e. assume that it establishes a given potential difference $V$ across its terminals irrespectively of the load, i.e. the total resistance around the circuit (this is the same as saying the power supply or battery has no "internal resistance"). Then the role of placing a series resistance in the circuit is to ensure that a fraction of the total voltage drop $V$ around the circuit occurs across the resistor, thereby reducing the voltage drop across the light bulb. It is the power dissipated in the latter (by Joule heating, the current $x$ voltage product), not the current per se, that determines whether it may overheat. The circuit shown (two purely resistive elements in series) is often known as a "voltage divider."

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