If the voltage applied to both ends of a fixed value resistor increases from 6V to 10V and the current through the resistor changes by 0.1A, the electric power of the resistor changes () A. 0.6WB. 1.6WC. 2WD. 3.4W

If the voltage applied to both ends of a fixed value resistor increases from 6V to 10V and the current through the resistor changes by 0.1A, the electric power of the resistor changes () A. 0.6WB. 1.6WC. 2WD. 3.4W

When the voltage at both ends of the resistance changes, the resistance remains unchanged. When the voltage at both ends of the resistance is 6V, the current through the resistance is I1 = 6vr, and the electric power of the resistance is P1 = (6V) 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 change △ I = i2-i1 = 10vr-6vr = 0.1A, so r = 40 Ω, so the electric power of the resistance changes It is shown that △ P = p2-p1 = (10V) 2R - (6V) 2R = (10V) 240 Ω - (6V) 240 Ω = 1.6W

In the experiment of measuring the electric power of small bulb, the power supply voltage is 4.5V, the rated voltage of small bulb is 2.5V, and the resistance is about 10 Ω
How to compare the rated power of a bulb with the same shape and rated voltage in series? How to compare the rated power of a bulb with the same shape and rated voltage in parallel?

The switch should be disconnected. If the ammeter is connected with a small range of 0.6A, you can't see the diagram. If it is connected to the top left and bottom right, you can slide to the leftmost end. If it is connected to the top left and bottom left, you can slide to the rightmost end. If it is connected to the top right and bottom right, you can slide to the leftmost end. The resistance of the sliding rheostat is the maximum!
Can it solve your problem?

In the experiment of "measuring the electric power of small light bulb", the rated voltage of small light bulb selected by Xiao Ming is 2.5V, the resistance is about 8 Ω, and the maximum resistance of sliding rheostat is 10 Ω. (1) in order to successfully complete the experiment, there are two battery packs with voltage of 4.5V and 6V for selection______ . A. only 4.5V battery pack can be selected & nbsp; & nbsp; & nbsp; & nbsp; B. only 6V battery pack can be selected C. both battery packs can be selected & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; D. It must be changed into a 10.5V battery pack. (2) please use a stroke line instead of a wire. According to the circuit diagram shown in figure a, connect the physical circuit in Figure B completely (the wires shall not be crossed, and the resistance of sliding rheostat connected into the circuit before closing the switch is the maximum). (3) move the slide of rheostat to make the voltmeter display 2.5V, and the current indication is shown in Figure C , whose value is______ A. The rated power of small bulb is______ W. (4) Xiaohui, Xiaoming's classmate, accidentally damaged the ammeter when she finished the experiment with the help of the above experimental equipment. Because there was no extra ammeter to replace, the teacher took the ammeter away and connected it to the original circuit with a resistance box instead of a sliding rheostat. ① after closing the switch, the light bulb did not light up, and the voltmeter's indication was equal to the power supply voltage, so the cause of the failure may be that Light bulb______ (select "short circuit" or "open circuit"); ② after troubleshooting, when the small bulb is working at the rated voltage, the resistance value of the circuit connected to the resistance box is R0, then the expression for calculating the rated power of the small bulb is______ (constant supply voltage)

(1) It can be seen from the circuit diagram that the bulb is connected in series with the sliding rheostat. When the bulb lights normally, the current in the circuit is about I = ulrl = 2.5v8 Ω = 516A. If the voltage of the power supply is u = 4.5V, the voltage at both ends of the sliding rheostat is u-ul = 4.5v-2.5v = 2V, and the resistance of the sliding rheostat connected to the circuit is r = u, I = 2v516a

The following is a student's experimental data for measuring the resistance and electric power of a small bulb with a rated voltage of 2.5V. Please fill in the missing contents:
Experiment times voltage value / v current value / a filament resistance / Ω bulb power / W
1 1 2.00 0.50 （ ）
2 2.5 0.24 （ ） （ ）
3 3.0 0.28 （ ） （ ）
According to the data in the table, try to find out;
(1) The resistance of the small bulb is () Ω
(2) The rated power of small bulb is () W
(3) What else do you find after analyzing the above data?

1 1 2.00 0.50 （ 2 ）
2 2.5 0.24 （10.4 ） （ 0.6 ）
3.3.0 0.28 （ 10.7 ） （0.84 ）
According to the data in the table, try to find out;
(1) The resistance of the small bulb is (7.2) Ω
(2) The rated power of small bulb is (0.6) W
(3) It is also found that the resistance of the filament increases with the increase of temperature;
If the voltage at both ends of the bulb is greater than the rated voltage, the power is greater than the rated power