If the voltage U and the resistance R are doubled simultaneously in a circuit, how will the power change? P = the square of (2U) divided by 2R Why is the square of (2R) divided by R

If the voltage U and the resistance R are doubled simultaneously in a circuit, how will the power change? P = the square of (2U) divided by 2R Why is the square of (2R) divided by R

Because P = u * I, I = u / R, so p = u ^ 2 / R. double each, that is p = (2U) ^ 2 / 2R

If the voltage applied to both ends of the constant value resistor increases from 8V to 10V, the current through the constant value resistor changes by 0.2A correspondingly, then the change of electric power consumed by the constant value resistor is ()
A. 0.4WB. 2.8WC. 3.2WD. 3.6W

When the voltage at both ends of the resistance changes, the resistance remains unchanged. When the voltage at both ends of the resistance is 8V, the current through the resistance is I1 = 8vr, and the electric power of the resistance is P1 = (8V) 2R. When the voltage at both ends of the resistance is 10V, the current through the resistance is I2 = 10vr, and the electric power of the resistance is P2 = (10V) 2R, and the current variation △ I

To connect a resistance wire with rated voltage of u to the circuit of U2 without changing its original power, you should ()
A. Connect a same resistance wire in series B. cut half of the resistance wire C. connect a same resistance wire in parallel D. connect the resistance wire into the circuit after folding it in half

A. When a same resistance wire is connected in series, the resistance becomes twice of the original, and the power of the resistance wire P1 = (U2) 22r = u28r = P8, so a is not in line with the title; B. cut half of the resistance wire, and the resistance becomes half of the original, and the power of the resistance wire P2 = (U2) 212r = u22r = P2, so B is not in line with the title; C. connect a same resistance wire in parallel, and the resistance becomes half of the original, and the resistance wire becomes half of the original The power of the resistance wire is P3 = P2, so C is not in line with the meaning of the problem; D. fold the resistance wire into the circuit, and the resistance becomes one fourth of the original. At this time, the power of the resistance wire is P4 = (U2) 214r = u2r = P, so D is in line with the meaning of the problem