In a series circuit, the higher the resistance, the higher the voltage?

In a series circuit, the higher the resistance, the higher the voltage?

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It is proved that there are two resistors R1 and R2 in series in the circuit with voltage u, and their powers are P1 and P2 respectively. If they are connected in parallel on the same circuit, the total power of the circuit is p = (P1 + P2) ^ 3 divided by p1p2

P1 = I string * I string * R1 (I string = u / (R1 + R2) = u * u * R1 / [(R1 + R2) * (R1 + R2)] --- formula 1P2 = u * u * R2 / [(R1 + R2) * (R1 + R2)] --- formula 2 substituting Formula 1, formula 2 then (P1 + P2) ^ 3 = {U * u * R1 / [(R1 + R2) * (R1 + R2)] + U * u * R2 / [(R1 + R2) * (R1 + R2)} ^ 3 = {U * u * (R1 + R2) / [(R1 + R2) * (R1 + R

It is proved that when the resistance is constant, the current is proportional to the voltage
The following materials are available: one power supply, one constant resistance, one sliding rheostat, one voltmeter, one ammeter, one switch, and several conductors. Please design an experiment to prove that when the conductor resistance is constant, the current through the conductor is proportional to the voltage at both ends of the conductor. Please draw the circuit diagram of the experiment and write the experimental steps

This is a physics exercise in middle school. It is an experiment to verify Ohm's law

Using Ohm's law to prove the quantitative relationship between resistance and voltage in series circuit

U1\U2=R1\R2
U1=I1*R1 U2=I2*R2
So U1 / U2 = I1 * R1 / I2 * R2
Because the current of series circuit is equal, i1.i2 is lost
U1\U2=R1\R2