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?

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?

Wow, the voltage drop of your ammeter is 1 ohm x 1 ampere = 1 V, not small
====According to your request:
To parallel a resistor
The shunt of this resistor (current through it) = 5A - 1A = 4A
Therefore, the resistance = u / I = 1V / 4A = 0.25 Ω
Power P = UI = 1 * 4 = 4W
Parallel a 0.25 ohm, 5W resistor
Ammeters are generally parallel resistors
U=I1*r=1A*1Ω=1V
R total = u / I2 = 1V / 5A = 0.2 Ω
1 / R total = 1 / R + 1 / R
1/0.2=1/R+1/1
R = 0.25 Ω
A 0.25 ohm resistor should be connected in parallel.
Given that the internal resistance of the ammeter is 0.1 ohm, and the range is 0 ~ 0.6, how many ohmic resistors do you need to connect to a 3-amp ammeter?
Calculation: the full scale voltage drop of the ammeter is 0.1 × 0.6 = 0.06 v. after extending the measuring range, the current flowing through the shunt resistor should be 3-0.6 = 2.4 A at full scale, then the resistance of the shunt resistor should be 0.06/2.4 = 0.025 ohm
3A/0.6A=5
0.1 ohm / (5-1) = 0.025 ohm
A resistance of 0.025 ohm is required in parallel
How many kilowatts can a square copper wire multiply and how much current can it multiply 2.54610.95 square meters
The diameter of copper wire at different temperatures and the maximum amperemeter it can bear 2008-04-28 06:22 p.m
The safe current carrying capacity of 2.5 square millimeter copper power line is - 28a
The safe current carrying capacity of 4mm2 copper power cord is 35A
The safe current carrying capacity of 6mm2 copper power line is 48A
The safe current carrying capacity of 10 square millimeter copper power line is - 65A
The safe current carrying capacity of 16 square millimeter copper power line is 91A
The safe current carrying capacity of 25mm2 copper power cord is - 120a
If it is aluminum wire, the wire diameter should be 1.5-2 times of copper wire
If the current of copper wire is less than 28a, it is safe to take 10A per square millimeter
If the current of copper wire is greater than 120a, the value is 5A per square millimeter
According to the total number of normal conducting current, the following can be selected:
Ten five, one hundred two, two five three five four three, seventy nine five two and a half, copper upgrade calculation
Let me explain to you that for aluminum wires less than 10 square meters, the number of square millimeters multiplied by 5 will be OK. For copper wires, it will be upgraded to a higher level. For example, for copper wires with 2.5 square meters, it will be calculated as 4 square meters. For those with more than 100 square meters, the cross-sectional area will be multiplied by 2. For those with less than 25 square meters, it will be multiplied by 4. For those with more than 35 square meters, it will be multiplied by 3,
Explanation: it can only be used as estimation, not very accurate
In addition, it is safe to keep in mind that the current per square meter of copper wire is less than 10A if the wire is less than 6 square millimeter. From this point of view, you can choose 1.5 square copper wire or 2.5 square aluminum wire
Within 10 meters, the current density of 6 A / sq.mm is more appropriate, 10-50 meters, 3 A / sq.mm, 50-200 meters, 2 A / sq.mm, and less than 1 A / sq.mm above 500 meters. From this point of view, if not far away, you can choose 4 square copper wire or 6 square aluminum wire
If it is 150 meters away from the power supply (not to say whether it is a tall building), it must use 4 square copper wire
The impedance of the wire is directly proportional to its length and inversely proportional to its diameter. When using the power supply, pay special attention to the wire and diameter of the input and output wires, so as to prevent accidents caused by overheating of the wire due to over current
The following is the copper wire diameter at different temperatures and the maximum current table
Wire diameter (approximate) (mm2)
Copper wire temperature (℃)
Sixty
Seventy-five
Eighty-five
Ninety
Current (a)
2．5
Twenty
Twenty
Twenty-five
Twenty-five
4．0
Twenty-five
Twenty-five
Thirty
Thirty
6．0
Thirty
Thirty-five
Forty
Forty
8．0
Forty
Fifty
Fifty-five
Fifty-five
Fourteen
Fifty-five
Sixty-five
Seventy
Seventy-five
Twenty-two
Seventy
Eighty-five
Ninety-five
Ninety-five
Thirty
Eighty-five
One hundred
One hundred
One hundred and ten
Thirty-eight
Ninety-five
One hundred and fifteen
One hundred and twenty-five
One hundred and thirty
Fifty
One hundred and ten
One hundred and thirty
One hundred and forty-five
One hundred and fifty
Sixty
One hundred and twenty-five
One hundred and fifty
One hundred and sixty-five
One hundred and seventy
Seventy
One hundred and forty-five
One hundred and seventy-five
One hundred and ninety
One hundred and ninety-five
Eighty
One hundred and sixty-five
Two hundred
Two hundred and fifteen
Two hundred and twenty-five
One hundred
One hundred and ninety-five
Two hundred and thirty
Two hundred and fifty
Two hundred and sixty
The wire diameter is generally calculated according to the following formula:
Copper wire: S = IL / 54.4 * u`
Aluminum wire: S = IL / 34 * u`
Where: I -- the maximum current passing through the conductor (a)
L -- length of conductor (m)
U '- allowable power drop (V)
S -- cross sectional area of conductor (mm2)
explain:
1. The voltage drop of U 'can be selected according to the voltage rating of power supply supplied by the equipment (such as detector) in the whole system
2. The calculated sectional area is upward. Estimation of current carrying capacity of insulated conductor
Multiple relationship between current carrying capacity and cross section of aluminum core insulated conductor
Conductor section (mm 2) 1 1.5 2.5 4 6 10 16 25 35 50 70 95 120
The current carrying capacity is 9 8 7 6 5 4 3.5 3 2.5
Current carrying capacity (a) 9 14 23 32 48 60 90 100 123 150 210 238 300
Estimation formula: 2.5 times 9, up minus 1 Shun number. 35 times 3.5, double group minus 1.5. The conditions are as follows: change plus conversion, high temperature 10% copper upgrade. The number of tubes is 234, 876% full load current. Note: (1) the formula in this section does not directly point out the current carrying capacity (safety current) of various insulated wires (rubber and plastic insulated wires), It can be seen from table 53 that the multiple decreases with the increase of the cross section. "2.5 times 9, up minus 1" means that the current carrying capacity of aluminum core insulated wires with various cross sections of 2.5mm 'and below is about 9 times of the number of cross sections, The current carrying capacity is 2.5 × 9 = 22.5 (a). The multiple relationship between the current carrying capacity and the number of cross-sections of the conductor from 4 mm 'and above is arranged upward along the wire number, and the multiple is reduced by 1, namely 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4. The current carrying capacity of the conductor from 35 mm' and above is 3.5 times of the number of cross-sections, that is, 35 × 3.5 = 122.5 (a), The multiple relationship between the current carrying capacity and the number of cross sections becomes a group of two wire numbers, and the multiple is reduced by 0.5 in turn. That is, the current carrying capacity of 50 and 70mm 'conductor is 3 times of the number of cross sections; the current carrying capacity of 95 and 120mm' conductor is 2.5 times of the number of cross sections, and so on, If the aluminum core insulated wire is exposed in the area where the ambient temperature is higher than 25 ℃ for a long time, the current carrying capacity of the wire can be calculated according to the above formula, and then 10% off. When the aluminum core insulated wire is used instead of the aluminum wire, its current carrying capacity is slightly larger than that of the aluminum wire of the same specification, The current carrying capacity of 16mm 'copper wire can be calculated as 25mm2 aluminum wire
Can a 6 kW water heater use 2.5 square wires
For the calculation formula of resistive load: P = UI, the current of 6000 watt electric appliance is equal to 6000 / 220 = 27a, and the general copper wire is calculated as 6A / mm2, then the maximum current of 6 square copper wire is 36a, and its maximum power is equal to 36a * 220V = 7920w. At the same time, the maximum current of 2.5 square copper wire is 15a, and its maximum power is only 15A * 220V = 3300W
I hope it will be useful to you
How many joules is 44 kwh?
44 * 1000 * 3600 = the eighth power of 1.584 * 10
1.584e8
44*1000*3600J=158400000J
All units are converted into international system of units.
The unit of kilowatt hour can be converted into 1000W * 3600s, which is equal to 3600000j
44 kwh = 158400000j
44 kwh = 158400000j~~
44kwh * 3.6 * 10 ^ 6 = 1.584 * 10 ^ 8 joules
44kw.h=44*3.6*10^6=158400000J=1.584*10^8~~~~~~~~~~~~~~~~~~~~(*^__ ^ *
1.584*10^8J=158400000J
Why is the delta connection voltage of this motor 220 V and the star connection 380 V?
 
The rated voltage of each phase winding of this motor is 220 v
If the input voltage of the motor is 380V (three-phase), then you use star connection. In this way, the voltage on each phase winding of the motor is 220V,
If your input voltage is 220 V (three-phase), then you use the triangle connection method. In this way, the voltage borne by each phase winding of the motor is also 220 v
If you input 380V voltage, according to the triangle connection method, the voltage applied on each phase winding is 380V, so the motor will burn
Maximum load of 2.5m2 copper core wire
National standard wire
1.5 square mm copper core wire can withstand 2200W load,
2.5 square mm copper core wire can bear 3500W load,
4mm2 copper wire can withstand 5200w load,
6mm2 copper wire can bear 8800w load,
10 square millimeter copper core wire can bear 14000W load
2. If the length ratio of two resistance wires of the same material and thickness is 2 ∶ 3, the resistance ratio is: ()
A、2∶3 B、3∶2 C、4∶9 D、9∶4
If the length ratio of two resistance wires of the same material and thickness is 2 ∶ 3, the resistance ratio is: (a, 2 ∶ 3)
The longer the length is, the greater the resistance is, so the longer the resistance wire is, the larger the resistance wire is, and the shorter the resistance wire is, the smaller the resistance wire should be
How many joules is 0.1 kwh?
And how many watts is 0.04 kilowatt?
0.04kW=40W
0.1kWh=100Wh=100*3600 J=360000J
6 × 10 quintic J
Why can 220 V / 380 V dual-purpose electric welding machine use both 220 V and 380 V? How can it be transformed
A: We specialize in the production of welding machines: the principle is very simple
Because the electric welding machine is just like a transformer, it has primary (power supply) and secondary (welding wire and ground wire). The transformer has an eternal formula, that is, primary voltage divided by the number of primary coils = secondary voltage divided by the number of secondary coils;
Therefore, when the primary is connected to 220 V, the number of coils is less, when it is connected to 380 V, the number of coils is more, so that the secondary voltage can remain unchanged and the welding current will remain unchanged. The transformer will not be burned