The resistance of a bulb is 440 ohm, which is connected to the lighting circuit, and the current through the bulb is 440 ohm____ A. Power on 1H, this bulb consumes a lot of energy___ Electric energy of J The formula and process should be used

The resistance of a bulb is 440 ohm, which is connected to the lighting circuit, and the current through the bulb is 440 ohm____ A. Power on 1H, this bulb consumes a lot of energy___ Electric energy of J The formula and process should be used

I=U/R=220/440=0.5A
W=U²/R=48400/440=110J

There are 20 bulbs with 484 ohm resistance. How to calculate the equivalent resistance after parallel connection
It's like 484 * 484 * 484 * 484 - divided by 484 + 484 + 484 + 484 - that's different
The formula I use is R1 * R2, R1 + R2, which is a pile of confusion
Why is the answer 484 divided by 20 equal to 24.2? What's the formula

Let n be the number of resistors and R0 be the resistance value
R total = R 0 / ln = 484 euro / 20 euro = 24.2 euro
Analysis: because the resistance parallel connection is equivalent to increasing the cross-sectional area of the conductor, when 20 resistors with the same resistance value are in parallel connection, it is equivalent to increasing the cross-sectional area of the original conductor by 20 times, You can deduce it according to the rule that the reciprocal of the total resistance of the parallel circuit is equal to the reciprocal sum of the partial resistances
I hope it can help you

When the following electrical appliances are working normally, it is necessary to prevent the current thermal effect ()
A. Electric light B. computer C. electric cooker D. electric iron

Electric lamp, electric cooker and electric iron mainly use the thermal effect of electric current to convert electric energy into internal energy; computer mainly converts electric energy into sound energy and light energy to prevent the heat generated by the thermal effect of electric current as far as possible

When the electric heater with rated voltage of 220 V works normally, the power consumption is 0.15 degree every 5 minutes
When the electric heater with rated voltage of 220 V works normally, the power consumption is 0.15 degree every 5 minutes. Question: (1) what is the rated power of the electric heater? (2) when the voltage at both ends drops to 200 V, how long does it take to get 9 × J heat from the electric heater? (set the resistance constant when the electric heater works)

（1）P=W/t=0.15kWh/（5/60）h=1.8kW=1800W.
(2) The resistance of electric heater is r = u & # 178 / P = 220 & # 178 / 1800 = 26.89 (Ω)
When the voltage drop is 200V, the actual power of the electric heater is: P '= u' &# 178 / r = 200 & # 178 / 26.89 = 1487.6 (W)
At present, W '= P' T1, T1 = w '/ P' = 9000j / 1487.6w = 6.05 (SEC)
The "9 × J" given in the title is wrong. The calculation is based on 9kj, that is 9000j. If it is other data, please replace 9000j in the last formula