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What can an ammeter measure besides the current?
The ammeter can not only measure the current, but also indirectly measure the power and circuit resistance. Of course, it needs simple calculation
When measuring the current with an ammeter, the correct way to do it is not to estimate the current in advance
A. First take a wire head of the circuit to try to touch the terminal with a larger range. If the pointer indication is within a smaller range, then use a small range
B. First take one end of the circuit to try to touch the terminal with a small range. If the pointer indicates a large number, then use a large range
C. Choose any range
D. Direct use of a large number of procedures
A
A
Is the accuracy of digital ammeter or ammeter better?
Both digital ammeter and ammeter can measure current. The problem is how accurate the current you want to measure. Choose the meter with corresponding accuracy according to the current accuracy you need to measure
In the circuit diagram shown in the figure, the indication of voltmeter V is 6 V, and that of ammeter A1 is 0.5 A. if the resistance of R2 is 6 ohm, calculate: (1) the indication of ammeter a I. (2) the total resistance of the circuit R
(1) The current through resistance R2: I2 = UR2 = 6v6 Ω = 0.1A, ∵ in parallel circuit, the main current is equal to the sum of branch currents ∵ the indication of ammeter A: I = 0.5A + 0.1A = 0.6A; (2) ∵ I = ur, ∵ the total resistance of the circuit: r = UI = 6v0.6a = 10 Ω. Answer: (1) the indication of ammeter A is 0.6A. (2) the total resistance of the circuit is 10 Ω
What is the parallel resistance between a 100 ohm 4 / 1 watt resistor and a 200 ohm 4 / 1 watt resistor, and the power becomes 2 / 1 watt
Resistance 66.7 Ω, power 0.5W
The resistance is 66.66 Ω and the total power should not exceed 0.375 W
No, they work at different voltages
There are five "220v800w" white burning lamps in a family, one with the same power resistance furnace. How to calculate the total power, total current, and electric energy consumed in a month?
Each electric furnace with 5 grid lamps has 800 watts, and the total power is 6 * 800 = 4800W = 4.8KW
Current = 4800 / 220 = 21.82a ≈ 22a
Monthly power consumption = 30 days * 24 hours / day * 4.8KW = 3456 degrees
Each electric furnace with 5 grid lamps has 800 watts, and the total power is 6 * 800 = 4800W = 4.8KW
Current = 4800 / 220 = 21.82a ≈ 22a
Monthly power consumption = 30 days * 24 hours / day * 4.8KW = 3456 degrees
800 * 5 = 4kw * 2 = 8kw
I=P/U
220/8=2.5A
8 * 24 * 30 = 5760 kw
It should be parallel
P lamp total = 800 * 5 = 4kw
P total = 2p total = 8kw
I = P divided by u = 36.4a
W = Pt = 8000w * 30 * 24 * 60 * 60 = 20726000000 joules
Two lamps marked with the words of 220v100w and 220v40w are connected in series in the circuit of 220 v. the ratio of power consumed by the two lamps is?
What is the reason
R = u & sup2 / P is obtained from P = UI and I = u / R, respectively
R1=U²/Po'=(220V)²/100W=484Ω,
R2=U²/Po''=(220V)²/40W=1210Ω；
P = I & sup2; R is obtained from P = UI and I = u / R, and the actual power ratio of two bulbs in series is calculated
P'/P''=I²R1/I²R2=R1/R2=484Ω/1210Ω=2/5
Series, the same current, the power ratio is equal to the resistance ratio, is equal to the inverse ratio of rated power, is 5:2
The current of the series circuit is equal. According to P = I ^ 2R, the power is proportional to the resistance. The resistance ratio of the two bulbs calculated by P = u ^ 2 / R is the inverse ratio of the power, so the power ratio is 2:5.
Formula P = u ^ 2 / R R1 = u ^ 2 / P = (220x220) / 100 = 22x22 R2 = u ^ 2 / P = (220x220) / 40 = 110x11
P = I ^ 2XR series current L1 / L2 = R1 / r2 = 22x22 / (110x11) = 2:5
Method: 100W / 220V = 0.45a, 40W / 220V = 0.18a
220 V / 0.45 = 488.8 Ω 220 V / 0.18 = 1222 Ω
220v/（488.8+1222）=0.128A
P=I*I*R I*I*R1/（I*I*R2）=P1/P2=R1/R2
=488.8/1222=0.4=2/5
Series, the same current, the power ratio is equal to the resistance ratio. Calculation by power formula with resistance
Series, the same current, the power ratio is equal to the resistance ratio, is equal to the inverse ratio of rated power, is 2:5, the answer is written in reverse.
The total power is 14 kW. How many meters should I buy
When single-phase power supply is used, select single-phase meter: its current is equal to = 14000 / 220 / 0.8 = 80A, in which the power factor is assumed to be 0.8; when three-phase power supply is used, select three-phase meter: its current is equal to = 14000 / 380 / 1.732/0.8 = 27a, in which the power factor is set to be 0.8, and 1.732 is the value of 3 root
Which is the same as the rated one in terms of actual current, voltage and power?
Suppose the load resistance value of an electrical appliance is a ohm, and when the voltage applied to it is b v, then the current flowing through it and the power consumed by it can be calculated. Do you think it is a fixed value? Or when the current applied to it is constant, then the voltage applied to it and the power consumed by it can be calculated, then do you think it is a fixed value? Besides, If it consumes a certain amount of power, is it a fixed value to calculate the voltage applied to it and the current flowing through it?
As long as there is one condition: this appliance is a fixed load, resistance load, by V = IR and P = UI, you can only know two values (one is the resistance value, the other is the voltage or current or power you applied), you can know the other two values, everything is a fixed value, brother!
It's all the same
What is the resistance of a 220 V, 800 w electric furnace in normal operation
P = UI = u & # / R (P --- power, u --- voltage, R --- resistance)
So r = u & # 178 / P = 220 & # 178; △ 800 = 60.5 Ω
5 Ω, P = u / R
sixty point five
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. If the indication of the ammeter increases by 4 times, R? Ohm resistance should be given Such as the title Better have an explanation
- 4. 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______ .
- 5. 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
- 6. 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．
- 7. 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
- 8. 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
- 9. 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
- 10. 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?
- 11. What do the poles and zeros of a filter mean
- 12. 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?
- 13. What do electric fish and volt battery mean
- 14. 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
- 15. 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?
- 16. 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
- 17. 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
- 18. 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?
- 19. How many joules is one degree of electricity
- 20. Physics of senior two! In the chapter of electromagnetic induction, we need to find Joule heat and use the conservation of energy. Why not calculate the work done by Ampere force?