In the circuit shown in the figure, the power supply voltage remains unchanged, R1 = 10 Ω. When the switch S is closed and the slide P of the sliding rheostat is in the middle point, the indication of the ammeter is 0.3A; when the slide P is moved to the rightmost end, the indication of the ammeter is 0.2A. Then the power supply voltage is 0.3A______ 5. The maximum resistance of sliding rheostat is______ Ω．

In the circuit shown in the figure, the power supply voltage remains unchanged, R1 = 10 Ω. When the switch S is closed and the slide P of the sliding rheostat is in the middle point, the indication of the ammeter is 0.3A; when the slide P is moved to the rightmost end, the indication of the ammeter is 0.2A. Then the power supply voltage is 0.3A______ 5. The maximum resistance of sliding rheostat is______ Ω．

If the voltage is u and the maximum resistance of the sliding rheostat is r, then when the slide is at the middle point, the indication of the ammeter (i.e. main circuit current) can be obtained by Ohm's Law: I1 = ur1 + R2. When the slide is moved to the right end, the indication of the ammeter (i.e. main circuit current) can be obtained by I2 = ur1 + R. substituting the data, the result is: u = 0.3A × (10 Ω + R2) -- - (1) u = 0.2A × (10 Ω + R) -- (2) the simultaneous solution is: u = 6V, R = 20 Ω So the answer is: 6, 20
In the circuit shown in the figure, the power supply voltage remains unchanged, R1 = 10 Ω. When the switch S is closed and the slide P of the sliding rheostat is in the middle point, the indication of the ammeter is 0.3A; when the slide P is moved to the rightmost end, the indication of the ammeter is 0.2A. Then the power supply voltage is 0.3A______ 5. The maximum resistance of sliding rheostat is______ Ω．
If the voltage is u and the maximum resistance of the sliding rheostat is r, then when the slide is at the middle point, the indication of the ammeter (i.e. main circuit current) can be obtained by Ohm's Law: I1 = ur1 + R2. When the slide is moved to the right end, the indication of the ammeter (i.e. main circuit current) can be obtained by I2 = ur1 + R. substituting the data, the result is: u = 0.3A × (10 Ω + R2) -- - (1) u = 0.2A × (10 Ω + R) -- (2) the simultaneous solution is: u = 6V, R = 20 Ω So the answer is: 6, 20
When RX (resistance to be measured) / RA (internal resistance of ammeter) is greater than RV (resistance of voltmeter) / Rx, the internal connection method of ammeter is adopted; if it is less than Rx, the external connection method is adopted?
When RX > (ra * RV) ^ 1 / 2, the internal connection method is used, Rx
Rx is much larger than RA
What are the models (rated current, voltage, power) of various LED lamp beads?
3528 20mA 3.0-3.6v 0.06w3014 30mA 3.2-3.6v 0.1w4014 30mA 3.2-3.6v 0.1w5050 60mA 3.2-3.6v 0.18w5630 150mA 3.2-3.6v 0.5W high power 350mA 3.2-3.6v 1W
Is the actual power of an electrical appliance equal to its rated power when it works at rated voltage
That's right
Your understanding is correct
How many megohms is one ohm
One megaohm is 1000000 ohm. That's the sixth power of 10, which is 1000 kiloohm
1 ohm = 0.000001 megohm = 0.001 kiloohm
Joule heat in physical circuits
Is the sum of Joule heat of R1 and R2 equal to the total Joule heat of R?
What is the relationship between Joule heat and resistance? Why is q = u ^ 2 / R inversely proportional and q = I ^ 2rt proportional?
Yes;
It is neither proportional nor inverse, because there is more than one variable in the relation. It can only be said that Q is inversely proportional or proportional to u or I when u or I are fixed
A bulb marked with "36V & nbsp; 40W" is connected to a circuit, and the measured current is 1a, then the actual voltage at both ends of the bulb is 1a______ 5. The actual power is______ W. (regardless of the influence of temperature on filament resistance)
∵ P = u2r, ∵ bulb resistance R = u2p = (36V) 240W = 32.4 Ω; ∵ I = ur, ∵ bulb actual voltage U = IR = 1a × 32.4 Ω = 32.4v, bulb actual power P = UI = 32.4v × 1A = 32.4w; so the answer is: 32.4; 32.4
Rated power and rated voltage of general household appliances
Household appliances like refrigerator, air conditioner, washing machine, etc
Name of electrical appliances general power (watt) estimated power consumption (KWH)
Window air conditioner 800-1300 maximum 0.8-1.3 per hour
Household refrigerator 65-130 about 0.85-1.7 per day
Household washing machine single cylinder 230 maximum 0.23 per hour
Double cylinder 380 maximum 0.38 Per Hour
Microwave oven 950
1500 0.16 per 10 minutes
0.25 every 10 minutes
Electric shower 1200
2000 per hour 1.2
2 per hour
Electric kettle 1200 1.2 per hour
Rice cooker 500 0.16 every 20 minutes
Electric iron 750 0.25 every 20 minutes
Hair dryer 450 0.04 every 5 minutes
Vacuum cleaner 400
850 0.1 every 15 minutes
0.21 every 15 minutes
Ceiling fan large 150 0.15 per hour
Small 75 0.08 per hour
Table fan 16 "66, 0.07 per hour
14 inch 52 per hour 0.05
TV 21 inch 70 0.07 per hour
25 inch 100 per hour 01
VCR 80 0.08 per hour
Audio equipment 100 0.1 per hour
Electric heating 1600-2000 maximum 1.6-2.0 per hour
The rated voltage is 220 V in China
The voltage is 380V, the power is 40kW, the resistivity is 1.7 * 10 ^ - 8 ohm · M. can you know how many square millimeter wires are needed?
I understand what you mean now. You should run a factory. You mean the safe load. Wait a moment, you can calculate one. According to the human body resistance and the maximum current limit, you can get that the safe load of copper wire is 5A ~ 8A / mm2, I = P / u = 105a, so the cross-sectional area of copper wire should be 13 ~ 21mm