When the voltage at both ends of the resistor increases from 2V to 4V, the current intensity through the resistor changes by 0.25A, Then the resistance of the constant resistance ro is____ Ohm. The electrical power of the resistance ro increases____ w. This is to fill in the blanks!

When the voltage at both ends of the resistor increases from 2V to 4V, the current intensity through the resistor changes by 0.25A, Then the resistance of the constant resistance ro is____ Ohm. The electrical power of the resistance ro increases____ w. This is to fill in the blanks!

U1=I*R
U2=(I+0.25)*R
2=I*R
4=(I+0.25)*R
Solution
I = 0.25 a
R = 8 Ω
P1 = U1 * I = 2 * 0.25 = 0.5W
P2 = U2 * (I + 0.25) = 4 * (0.25 + 0.25) = 2W
Added p2-p1 = 2-0.5 = 1.5W

When the voltage applied at both ends of a constant value resistor is 4V, the current passing through the resistor is 0.5A,
If the voltage is changed to 6V, can a 0.6A ammeter be used to measure the current passing through the resistor?

Answer: No
Analysis:
R＝U/I＝4V/0.5A＝8Ω
When u = 6V:
I＝6V/8Ω＝0.75A＞0.6A
Therefore, it cannot be measured with an ammeter with a measuring range of 0.6A
I don't know what to ask

Physics work in grade 3: when the voltage at both ends of the resistor is 4V, the current passing through is 0.5A. If the voltage at both ends of the resistor increases to 5V, can we use an ammeter with a range of 0-0.6a to measure the current?

1. When the voltage at both ends is 4V, the current passing through is 0.5A, i.e. r = u / I = 4 / 0.5 = 8 Ω
2. When the voltage at both ends of the resistance increases to 5 V, the current I = u / r = 5 / 8 = 0.625a, 0.625a > the maximum limit of ammeter is 0.6A, so it cannot be measured