When the voltage at both ends of a fixed value resistor increases from 4 V to 6 V and the current through it increases by 0.4 a, how many watts does the electric power increase? When the voltage at both ends of the resistor is 6 V and the power on time is 5 minutes, how much heat is generated?

When the voltage at both ends of a fixed value resistor increases from 4 V to 6 V and the current through it increases by 0.4 a, how many watts does the electric power increase? When the voltage at both ends of the resistor is 6 V and the power on time is 5 minutes, how much heat is generated?

This kind of problem generally has two states. We should grasp the basic solution idea that the resistance does not change and the power supply voltage does not change
Let the constant resistance be r, and the current flow is I1 when the voltage is 4V
According to Ohm's law, if r = U1 / I1, then 4 / I = 6 / (I + 0.4), then I1 = 0.8A, then r = U1 / I1 = 4 / 0.8 = 5 ohm
Electric power increased: p2-p1 = U2 ^ 2 / R - U1 ^ 2 / r = 6x6 / 5-4x4 / 5 = 4W (^ 2 for square, P = UI = I ^ 2R = u ^ 2 / R)
Heat generated by power on for 5 minutes: q = U2 ^ 2T / r = 6x60x5 / 5 = 2160 joules (q = Pt = uit = I ^ 2rt = u ^ 2T / R)

When the voltage at both ends of the circuit is 6 V, the current through the resistance is 0.6 A. ask (1) how many V is the voltage at both ends of the resistance when the current through the resistance is 1 a?
(2) If the voltage at both ends of the circuit increases to 12 V, what is the resistance in the circuit and what is the current through the resistance?
From the question
（1）U,=IR=I *（U/I,）=1A*（6V/0.6A）=10V
(2) 10 Ω I = u / r = 12V / 10 Ω = 1.2A
Why is the answer to the second question 10 euro,

R = u / I, so r = 6 / 0.6 = 10 ohm;
U = R * I, so u = 1 × 10 = 10 Ford;
In the circuit, the size of the general resistance is fixed, so the size of the resistance is 10 ohm
That is, r = 10. I = u / R, so I = 12 / 10 = 1. 2 A;