The higher the rated power, the lower the resistance, the more heat per unit time of the bulb with the same rated voltage. However, according to Joule's law, the opposite is true,

The higher the rated power, the lower the resistance, the more heat per unit time of the bulb with the same rated voltage. However, according to Joule's law, the opposite is true,

Joule's Law: the heat generated by the current passing through the conductor is proportional to the square of the current intensity, the resistance of the conductor and the power on time. That is, q = I ^ 2rt can also be converted into q = uit, q = u ^ 2 / R * t. your bulb has the same rated voltage and different rated power. If the power is large, the current will be large. The square of the current is smaller than the resistance

The current of a 5 Ω resistor passing through it is 2a, and the heat released in 10 seconds is 2______ J. If the current through it is reduced to the original 12, the heat released in the same time is the original______ &Set the resistance value unchanged

Q = i2rt = (2a) 2 × 5 Ω × 10s = 200J; the current through it is reduced to the original 12: Q ′ = (I ′) 2rt = (12 × I) 2 × R × t = 14i2rt = 14q

Calculation problem: the resistance of a conductor is 10 ohm. Try to find the Joule heat produced by the conductor in one minute when 0.1 a current passes through

1min=60s .
W = uit = I & sup2; RT = (0.1A) & sup2; × 10 Ω × 60s = 6J

As shown in the figure, after the resistor R with a resistance value of 10 Ω is connected in series with the electric lamp L with a rated voltage of 3.8V, it is connected to both ends of the 4V power supply, close the switch s, and the measured electric power of the electric lamp L is 0.4W, then the voltage at both ends of the resistor R is_ 2_ 5. The current in the circuit is_ 0.2_ A. The rated power of lamp L is_ 1.444_ W (regardless of the effect of temperature on resistance)

Set resistance voltage as U1 and lamp voltage as U2
Because of series connection, the current is equal, so U1 / 10 = 0.4/u2 (formula I = u / R; u = P / I)
And because U1 + U2 = 4V
So we get UI = U2 = 2V
So I = 0.2A
Lamp resistance R = U2 / P = 4 / 0.4 = 10
So the rated power is p = U2 / r = 3.8 times 3.8 / 10 = 1.444