Why does the current always remain the same in a series circuit? When the charges encounter a resistor, wouldn't the speed of flow decrease? If the speed decreases, shouldn't the rate of flow decrease at the same time?
The first part of this question seems to be misleading and is thus confusing the answer.
A resistor does not inherently "decrease electric current" it actually resists the flow of current and causes a drop in voltage not the current.
I am sure you are aware of Ohm's law: V=I⋅RV=I⋅R or I=VRI=VR
So if you have a simple circuit with a constant voltage supply, increasing an adjustable resistance will reduce the current flowing through the "whole" simple circuit.
However, if you have a simple circuit with aconstant current source, by definition the current through the circuit will not change regardless of what the load resistance is. Increasing the resistance will increase the voltage drop across the resistor.
So, the premise in the first part of the question is not absolute.
If resistors are connected in series, then by Kirchhoff's circuit law, ∑na=1Ia=0∑a=1nIa=0, the current through both resistors will be the same since there is no other exit (no third branch I3I3) for the current between the resistors to exit. There is the input branch with a current flowing in (+I1+I1 from R1), and an output branch with the current flowing out (−I2−I2 to R2).
With Kirchoff's Law, the current through R1 (I1I1) must equal the current through R2 (I2I2) and in this simple circuit, the "current is [the] same in [the] whole circuit".
The first part of this question seems to be misleading and is thus confusing the answer.
A resistor does not inherently "decrease electric current" it actually resists the flow of current and causes a drop in voltage not the current.
I am sure you are aware of Ohm's law: V=I⋅RV=I⋅R or I=VRI=VR
So if you have a simple circuit with a constant voltage supply, increasing an adjustable resistance will reduce the current flowing through the "whole" simple circuit.
However, if you have a simple circuit with aconstant current source, by definition the current through the circuit will not change regardless of what the load resistance is. Increasing the resistance will increase the voltage drop across the resistor.
So, the premise in the first part of the question is not absolute.
If resistors are connected in series, then by Kirchhoff's circuit law, ∑na=1Ia=0∑a=1nIa=0, the current through both resistors will be the same since there is no other exit (no third branch I3I3) for the current between the resistors to exit. There is the input branch with a current flowing in (+I1+I1 from R1), and an output branch with the current flowing out (−I2−I2 to R2).
With Kirchoff's Law, the current through R1 (I1I1) must equal the current through R2 (I2I2) and in this simple circuit, the "current is [the] same in [the] whole circuit".
