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In the circuit shown in the figure, both ammeter and voltmeter are ideal ammeter. If resistance R1 is open circuit, the following statements are correct () A. The current indicates that the number becomes smaller; B. the voltage indicates that the number becomes larger; C. the power consumed by R3 becomes larger; D. the power consumed by the circuit in the power supply becomes larger
A. B. when resistance R1 is open circuit, the total resistance of the external circuit increases. According to Ohm's law of closed circuit, the total current I in the circuit decreases and the terminal voltage U increases. According to Ohm's law, the current of the ammeter increases and the voltage at both ends of resistance R2 increases, and the indication of the ammeter and the voltmeter increases. Therefore, a is wrong, B is correct. C. resistance R1 is open circuit, and the above analysis shows that it is on If the current of the over-current meter increases, the power consumed by R3 will increase. So C is correct. D, the total current in the circuit decreases. According to P = I2R, the power consumed by the circuit in the power supply will decrease. So D is wrong. So select BC
As shown in the figure, the maximum resistance of the rheostat is 100 Ω, and the resistance of the bulb remains unchanged. When the switch is closed and the slide P is at the a end, the indication of the ammeter is that the slide P is at the B end
Figure: a switch, power supply, bulb, ammeter, sliding rheostat in series
It can be seen from the meaning of the title that when the slide is at the a end, the resistance value of the rheostat access circuit is 0, while when the slide is at the B end, the resistance value of the rheostat access circuit is the largest, that is, 100 Ω
Let the power supply voltage be u and the resistance of the lamp be r
When the slider is at the a end, the current is I1 = u / R
When the slide is at the B end, the current is I2 = u / (r Lamp + R), and R is equal to the maximum resistance of the rheostat
So I 1 / I 2 = 5
U / R lamp = 5 * U / (r Lamp + R)
That is, R lamp = R / 4 = 100 / 4 = 25 Ω
Give you a battery pack, an ammeter, a constant resistance R0 with known resistance, a switch and several wires. Use these devices to measure a resistance RX with unknown resistance
Circuit: connect R0 and the resistance to be measured in parallel, the ammeter is installed on the main circuit, the switch is installed on the branch of the resistance to be measured, and then connect the power supply. Steps: 1. Connect to the power supply, the indication of the ammeter is i12. Close the switch, the indication of the ammeter is I2
As shown in the figure, the key K is closed, and the ammeter and voltmeter are ideal ammeters. If the resistance R1 is open circuit, the following statement is correct ()
A. The current indicates that the number becomes larger; B. the voltage indicates that the number becomes smaller; C. the power consumed by R1 becomes larger; D. the power consumed by the circuit in the power supply becomes larger
A. According to Ohm's law of closed circuit, the total current I in the circuit decreases and the terminal voltage U increases. According to Ohm's law, the current of ammeter increases and the voltage at both ends of resistor R2 increases, and the indication of ammeter and voltmeter increases. Therefore, a is correct and B is wrong. C. the breaking of resistor R1 no longer consumes power D. the total current in the circuit decreases. According to P = I2R, the power consumed by the circuit in the power supply becomes smaller. Therefore, D is wrong. Therefore, a is selected
Two lamps marked with the words of 220 V 100 W and 220 V 40 W are connected in series in the 220 V circuit, and the power consumption ratio of the two lamps is______ ,
Why 2:5 instead of 5:2
When the two bulbs marked with the words of 220 V 100 W and 220 V 40 W work normally: P = u * U / r r r = u * U / P R1 = u * U / P1 R2 = u * U / P2, so: R1 / r2 = P2 / P1 = 40 / 100; after series voltage division: U1 / U2 = R2 / r2 P1 = I series * U1 P2 = I series * U2 P1 / P2 = U1 / U2 = R1 / r2 = 40 / 100, they are connected in series with 220 V power supply
The total power is 14 kW. How many meters should I buy
What's the maximum size of 220 V
The total power is 14kw, the maximum current is about 70A, buy 40-80a (long-term operation 40a, maximum current 100a)
The voltage is determined by______ The function of power supply is to provide voltage to both ends of electrical appliances; electrical appliances need to consume electric energy,
Power is also provided______ The device
The voltage is supplied by the power supply
The power supply is also a device that provides voltage
A "220 V 800 W" electric furnace, how big is the resistance of the resistance wire during normal operation? If the voltage in the circuit is reduced to 200 V, and the resistance of the resistance wire remains unchanged, how big is the actual electric power of the electric furnace?
Current: I = P △ u = 800W △ 220V = 40 / 11 (a)
Then resistance: r = u △ I = 220 V △ 40 / 11a = 60.5 Ω
P1=U×U/R≈661.16(W）
A: in normal operation, the resistance is 60.5 Ω, and the power is 661.16w,
When two lamps marked with the words of 220 V 100 W and 220 V 40 W are connected in series in a 220 V circuit, the ratio of power consumed by the two lamps is?
1. According to the electric power calculation formula: P = I & # 178; R
P1∶P2=I1²R1∶I2²R2
Because in the series circuit, the current through each resistor is equal, there is: I1 = I2; so P1 ∶ P2 = R1 ∶ R2
2. According to the formula: r = u & # / P
Constant value resistance of 100W bulb: R1 = 220 & #178 / 100 = 484 (Ω)
Constant resistance of 40W bulb: R2 = 220 & # 178 / 40 = 1210 (Ω)
3、P1∶P2=R1∶R2=484∶1210=2∶5
Answer: in the 220 V series circuit, the power consumption ratio of 100 W bulb and 40 W bulb is 2 ∶ 5
It can be seen that in the series circuit, the higher the rated power, the less power the bulb consumes, and the lower the power the bulb is brighter than the larger one
How to calculate power of power circuit
Power circuit is calculated by apparent power, that is: current = supply voltage / load impedance; apparent power = supply voltage × current
Because regardless of the power factor (which depends on the phase relationship between the current and voltage on the load), the maximum current flowing in the circuit is the real current. The so-called imaginary value is just a way to distinguish the current on the resistance, Virtual current can also burn lines or equipment. Therefore, the power of power circuit (lines or equipment) should be calculated according to current, not useful power
But the power consumed by the power circuit is still calculated as useful power, that is, apparent power × power factor, because the virtual power will eventually be fed back to the power supply, which is not the consumed energy
The power of three-phase electricity is the sum of the power consumed by the three phase lines, regardless of whether the current on the three phase lines is balanced or not (unbalanced will increase the extra energy loss of the line and equipment). Unbalanced three-phase electricity only increases the transmission loss of the line and equipment, and does not affect the calculation relationship of power, but only multiplies the power of one line by three when balanced, When unbalanced, the power of the three lines must be added up
A: the power balance factor of 3x80
RELATED INFORMATIONS
- 1. Measure the resistance with missing meter method, from the positive pole to the resistance RO and then to Rx, ammeter, switch to the negative pole, RX and RO are connected in parallel, and Rx resistance value is calculated with known power supply voltage ro resistance value
- 2. If the indication of the ammeter increases by 4 times, R? Ohm resistance should be given Such as the title Better have an explanation
- 3. As shown in the figure, given the resistance R1 = 3 Ω, R2 = 6 Ω, the indication of ammeter A is 0.6A, then the indication of ammeter A1 is 0.6A______ .
- 4. A circuit composed of a resistor, a power supply and an ammeter in series, The indication of the ammeter is 0.12A; if a 10 ohm resistor is connected in series in this circuit, the indication of the ammeter is 0.1A. Calculate the voltage of the power supply and the resistance in the original circuit
- 5. 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．
- 6. If the internal resistance Ra of the ammeter is 0.03 ohm and the internal resistance RV of the voltmeter is 2 kiloohm, the resistance R to be measured is about 1.5 kiloohm, and the internal and external connection methods have small error
- 7. The internal resistances of small range ammeter g, ammeter A and voltmeter V are RG, RA and RV respectively A. RG＜RA＜RVB. RV＜RA＜RGC. RA＜RG＜RVD. RG＞RV＞RA
- 8. The voltage at both ends of a conductor is 12 volts, and the current intensity passing through it is 0.9 amperes, so that the current intensity passing through it is 0.6 amperes, and the voltage applied at both ends of it should be more than 10 amperes
- 9. There is an ammeter with an internal resistance of 1 ohm and a measuring range of 1 A. if you want to expand the measuring range to 5 A, what kind of resistance should be connected? Is it in series or in parallel?
- 10. How many kilowatts of electrical appliances can 4 square millimeter copper wire use
- 11. What can an ammeter measure besides the current?
- 12. What do the poles and zeros of a filter mean
- 13. The physical layer is the foundation of all networks. How many volts are used to represent 1, how many volts are used to represent 0, and how long does a bit last for microseconds?
- 14. What do electric fish and volt battery mean
- 15. In the study of microscopic particles, electron volt (EV) is often used as the unit of energy. 1eV is equal to the kinetic energy of an electron accelerated by 1V voltage. Please deduce the conversion relationship between electron volt and Joule
- 16. In the motor model, the coil resistance is 1 ohm, the voltage at both ends is u = 2 V, and the current in normal operation is 0.8 ampere 1. Electric energy consumed in 1min when small motor works normally 2. The internal energy produced in 1min when the small motor works normally? 3. The mechanical energy produced in 1min when the small motor works normally? 4. Heating power of small motor in normal operation? 5. Total power of small motor in normal operation? 6. Output power of small motor in normal operation?
- 17. When the coil resistance of a motor is 1 ohm, the voltage applied at both ends of the coil is 2V, and the current is 0.8A, the electric energy consumed by the motor in normal operation for one minute is____ J, What is the converted mechanical energy J
- 18. When a motor works normally, the coil voltage is 220V, the coil resistance is 2 ohm, the current is 10a1s, the electric energy consumed is j, and the heat generated in 1min is J
- 19. The resistance of a motor coil is 1 ohm, When the voltage applied at both ends of the coil is 2V, the current is 0.8A, and the motor works normally. Calculate the electric energy consumed by the motor in normal operation for 1min The second question is how much mechanical energy is converted by the motor working for one minute?
- 20. How many joules is one degree of electricity