It is proved that the total resistance is equal to the sum of the resistances in series circuit

It is proved that the total resistance is equal to the sum of the resistances in series circuit

It is proved that the current in series circuit is equal everywhere (axiom), r = u / I
The power supply voltage is equal to the sum of the voltages of each section (axiom),
R total = (U1 + U2 +...) +Un）/I=U1/I+U2/I+…… +Un/I=R1+R2+…… +Rn
The process is a bit of crap, but that's the truth

Is the sum of the two branch currents of the parallel circuit multiplied by the sum of the two parallel resistors equal to the supply voltage

Equal to the terminal voltage of the two parallel branches
That is: if the power supply only supplies power for the two parallel circuits, and the internal resistance of the power supply is ignored, the multiplication result is the power supply voltage

In parallel circuit, when the resistance increases, how does the current change? How does the voltage change?
In a series circuit, when the resistance increases, how does the current change? How does the voltage change?
No, don't talk nonsense,

In the case of neglecting the internal resistance, in the parallel circuit, the resistance becomes larger, the current becomes smaller, and the voltage remains unchanged. Because the voltage at both ends of the parallel circuit is equal to the supply voltage. In the series circuit, the resistance becomes larger, the current becomes smaller, and the voltage becomes larger. Because the current becomes smaller, the voltage drop produced by other series resistors becomes smaller, so the voltage with larger resistance is increased