Electronic – Basics physics question about calculating resistance

Of course there is, but it does not look pretty. Make the divisors equal, and add the terms. for three resistors you get

\$ \dfrac{R2 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R2}{R1 \cdot R2 \cdot R3}\$

\$ = \dfrac{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)}{R1 \cdot R2 \cdot R3} \$

now do the \$ \dfrac{1}{n} \$ and you get:

\$ \dfrac{R1 \cdot R2 \cdot R3}{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)} \$

The top line is easy, it is the product (multiplication) of all resistors. The bottom line is the sum of the products of all leave-one-out combinations. For two that reduces to the pretty formula:

\$ \dfrac{(R1 \cdot R2)}{(R1+R2)}\$

how does the first resistor's resistance influences the second resistor's resistance.

It doesn't. However, the resistance of the first resistor influences the voltage across the second resistor.

Clearly, the resistors in the diagram are series connected thus the current through each resistor is identical.

$$I_1 = I_2 = I$$

By Ohm's law, we have

$$V_1 = I_1 \cdot R_1 = I \cdot R_1$$

$$V_2 = I_2 \cdot R_2 = I \cdot R_2$$

By KVL, we have

$$6V = V_1 + V_2 = I \cdot R_1 + I \cdot R_2 = I \cdot (R_1 + R_2)$$

Thus

$$I = \frac{6V}{R_1 + R_2}$$

In other words, the current \$I\$ depends on the sum of the resistances (series connected resistances add).

The voltage across the second resistor can now be written as

$$V_2 = I \cdot R_2 = \frac{6V}{R_1 + R_2} \cdot R_2 = 6V \frac{R_2}{R_1 + R_2} $$

and so, as first stated, the resistance of the first resistor influences the voltage across the second resistor.

Similarly

$$V_1 = I \cdot R_1 = \frac{6V}{R_1 + R_2} \cdot R_1 = 6V \frac{R_1}{R_1 + R_2}$$