The resistance of an electric furnace in normal operation is 80 ohm, which is connected to the lighting circuit and releases 72600 joules of heat. The working time of the electric furnace is______ ．

The resistance of an electric furnace in normal operation is 80 ohm, which is connected to the lighting circuit and releases 72600 joules of heat. The working time of the electric furnace is______ ．

The working time of electric furnace t = qru2 = 72600j × 80 Ω (220V) 2 = 120s = 2min

As shown in the figure, the power supply circuit remains unchanged, the bulb L is marked with "6V 2W", the resistance value of setting resistor R1 is 12 ohm, and the maximum resistance value of sliding rheostat R is 200 ohm
If the power supply voltage of a '220 V 300 W' electric iron fluctuates within 10%, the actual power of the iron will vary from - to -. If you can, can you write the idea

P = u * I, r = u / I, u ^ 2 / r = P can be calculated. R is not only the resistance of the iron, but also p = u * I = u ^ 2 / R. it can be seen that the corresponding ratio relationship between the power and voltage of the resistance is quadratic. If the voltage changes within 10%, the power changes between 8.11% and 121% of the rated power

The resistor R and the bulb L are connected in series to a circuit with a voltage of 10V, r = 10 Ω. After the circuit is connected, the work done by the current on the resistor R is 10J within 100s. It is known that the rated power of the bulb L is 10W, and the resistance of the bulb does not change
Using known forms to find solutions

Known: u = 10V, r = 10 Ω, t = 100s, w = 10J, PL = 10W
From w = I & # 178; RT, the current through the resistance is I = √ (w / RT) = √ (10 / 10 × 100) = 0.1 (a)
The resistance is connected in series with the bulb L, so the current flowing through the bulb is I = 0.1A. If the resistance of the bulb is RL, then:
I × (R + RL) = u, 0.1 × (10 + RL) = 10, the solution is RL = 90 (Ω)
From PL = UL & # 178 / RL, so: UL = √ (PL × RL) = √ (10 × 90) = 30 (V)
A: the rated voltage of the bulb is 30V

The resistor and the lamp are connected in series to a circuit with a voltage of 10V, r = 10. The work done by the current to the resistor R within 100s after the circuit is connected is 10J. The rated power of the lamp is known to be 10 lamp voltage

W=P*T=(U^2/R)*T=(U^2/10)*100=10
So u = 1V
So lamp voltage = 10-1 = 9V