Hello!

Let us speak about two resistors of the resistances Q and S. For three or more resistors the cause is the same.

By the definition, resistance of an electrical circuit is the voltage applied divided by the resulting current. It remains the same for different voltages (the current changes accordingly). In other words, the voltage drop on a resistor is IR, where I is the current and R is the resistance.

For two resistors connected in series, the same current I flows through one resistor and then through the second. The drop of voltage occurs twice: IQ and then IS. The resulting resistance is (IQ+IS)/I=Q+S which is greater than both Q and S.

For two resistors connected in parallel, the current divides on two parts, `I_1` and `I_2` , `I_1+I_2=I` . The endpoints of the resistors are connected, therefore the voltage drop must be the same, `I_1Q=I_2S.` It is simple to solve this linear system and obtain `I_2S=I S/(1+S/Q),` so the total resistance is `S/(1+S/Q)=1/(1/S+1/Q)` , which is smaller than both S and Q.

Less formally, series connection requires to push current through both barriers. Parallel connection allows current to pass with smaller parts through one resistor each part.

**Further Reading**

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