When an unknown resistance is connected to the power supply in series with a 6 Ω resistance, its voltage at both ends is 4V; when it is connected to the same power supply in series with a 9 Ω resistance, its voltage at both ends is 3V

Series of equations
U*R/(R+6Ω)=4V
U*R/(R+9Ω)=3V
The comparison of the two formulas gives r = 3 Ω and u = 12V

Two resistors R1 and R2 are connected in series to a 6V power supply, and the measured voltage at both ends of R1 is 4V. If they are connected in series to a 6V power supply, and the measured trunk current is 1.5A, then the resistance R1 = - R2 =?

In series
Voltage of R1 = 6 / (R1 + R2) * R1 = 4V
parallel connection
Trunk current = 6 / [R1 * R2 / (R1 + R2)} = 1.5
The solution is R1 = 12 ohm, R2 = 6 ohm
Because I guess you write a wrong word, if the future "series" I guess is parallel, so the solution is different from them

The voltage at both ends of a conductor is 4V, and the current passing through the conductor is 0.2A. (1) the resistance of the conductor is_ Ω
(2) If the current through the conductor becomes 0.3 A, the resistance of the conductor is_ Europe

(1) The conductor resistance is r = u / I = 4 / 0.2 = 20 Ω,
(2) When the current is 0.3A, the resistance R = 20 ohm

When the voltage at both ends of a section of conductor is 2V, the current in the conductor is 0.5 A, and the resistance of the conductor is____ If both ends of the conductor drop to 0V, the current of the conductor is? And the resistance is? If it increases to 4V, calculate the current

Resistance = 2V / 0.5A = 4ohm
When the voltage at both ends of the conductor drops to 0V, the current of the conductor is 0A and the resistance is 4ohm
When the voltage at both ends of the conductor increases to 4V, the current = 4V / 4ohm = 1a

When an unknown resistance is connected to the power supply in series with a 6 Ω resistance, its voltage at both ends is 4V; when it is connected to the same power supply in series with a 9 Ω resistance, its voltage at both ends is 3V