The maximum current that can be measured by an ammeter is 10mA. When the voltage at both ends of a resistor is known to be 8V, the current passing through it is 2mA. If a voltage of 50V is added to the resistor, can the ammeter be used to measure the current passing through the resistor? Please give your reason by calculation

According to Ohm's law, the resistance of constant resistance is r = u1i1 = 80.002 = 4000 Ω. When the voltage is 50V, the current passing through the constant resistance is I2 = u2r = 504000 = 0.0125a = 12.5ma ＞ 10mA. Therefore, the current passing through this resistance cannot be measured with this ammeter. A: it cannot be measured with this ammeter The current through this resistor
We should connect a "220 V, 100 W" light bulb to 380 V power supply temporarily to make the light shine normally? What is the electrical power consumed by this resistor?
∵ P = u2r, ∵ the resistance of normal light-emitting bulb is RL = ul2pl = (220V) 2100W = 484 Ω, ∵ P = UI, ∵ the current of normal light-emitting bulb is I = plul = 100w220v = 511A; ∵ I = ur, ∵ the total resistance of circuit is r = UI = 380v511a = 858 Ω, the resistance to be connected in series is R0 = r-rl = 858 Ω - 484 Ω = 374
How much power can 2.5 square multi strand copper wire bear
How many watts can 2.5 square multi strand copper wire bear under the condition of 220 V voltage
For example, if a light is 40W, how many lights should it be able to connect? Yes, there is no calculation formula
The partial resistance increases the full voltage of the original ammeter?
Why don't I change the voltmeter into a voltmeter in series?
For an ammeter, its full bias current Ig and internal resistance Rg are fixed and can no longer be changed. Similarly, the maximum voltage that it can withstand UG = Ig * RG is a fixed value. When the ammeter is connected with a voltage divider in series, the ammeter A and R should be regarded as a whole, that is, the ammeter v. now the full voltage of the ammeter V increases, not the full voltage of the ammeter a
How to calculate the current of a bulb with a voltage of 220 V and a voltage of 100 W? How many ohm is 100 W? How many ohm is 100 W? How many ohm is 100 W?
The power is equal to the voltage times the current. A 100W bulb consumes 0.45a
The power is equal to the square of voltage divided by resistance, and the resistance of 100W bulb is 484 Ω
The resistance of the bulb can be divided into cold state and hot state. The hot state resistance of 100W bulb is 484 Ω, and the cold state resistance is smaller than the hot state resistance
How much current can 120 square meters of copper wire bear
Such as the title
P=1.732UIX0.8
58a
The general calculation method of copper wire safety is as follows:
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
Estimation formula:
Multiply it by nine at 2.5 and go up minus one
Multiply 35 by 3.5, double double group minus 0.5
Conditions are variable plus conversion, high temperature 10 fold copper upgrade
The number of pipes is 234, 876
explain:
(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), but "cross section multiplied by a certain multiple", which is obtained by mental calculation. It can be seen from table 53 that the multiple decreases with the increase of cross section
"2.5 mm, multiply by 9, subtract one from the top" means that the current carrying capacity of aluminum core insulated wires with various cross sections of 2.5 mm 'and below is about 9 times of the number of cross sections. For example, the current carrying capacity of 2.5 mm' conductor is 2.5 × 9 = 22.5 (a). The multiple relationship between the current carrying capacity and the number of cross sections of 4 mm 'and above is arranged along the line number, and the multiple is successively reduced by 1, that is 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4
In other words, the relationship between the section of two conductors and the section of two conductors becomes 355.3 mm, That is to say, 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 not used, but the copper core insulated wire is used, Its current carrying capacity is slightly larger than that of aluminum wire of the same specification. According to the above formula, the current carrying capacity of one wire number larger than that of aluminum wire can be calculated. For example, the current carrying capacity of 16mm 'copper wire can be calculated as 25mm2 aluminum wire
The voltage at both ends of resistance R is 32V, and the current passing through is 0.4A. If the voltage increases to 56V, can we use an ammeter with two values of 0-0.a to measure it
What is the current of this resistor?
From Ohm's law
R = u / I = 32 / 0.4 = 80 Ω
When the voltage increases to 56V,
I‘=U’/R=56/80=0.7A.
The range of ammeter is 1.0A
The resistance value is 32 △ 0.4 = 80 Ω
When the voltage increases to 56V, the current is 56 △ 80 = 0.7A
You can't see the measurement range clearly, but it doesn't matter. As long as the measurement range exceeds 0.7A, it can be measured. Resistor R1 and R2 are connected in series in the circuit. It is known that the voltage at both ends of R1 is 3V, the voltage at both ends of R2 is 9V, and R1 + R2 = 400 Ω. Calculate the resistance value of R1 and R2. If they are connected in series, the current is equal. Therefore, the ratio of resistance value and voltage of the two resistors is the same, which is 3:9 = 1:3. And R1 + R2 =... Expansion
The resistance value is 32 △ 0.4 = 80 Ω
When the voltage increases to 56V, the current is 56 △ 80 = 0.7A
You can't see the measurement range clearly, but it doesn't matter. As long as the measurement range exceeds 0.7A, it can be measured. Inquiry: resistor R1 and R2 are connected in series in the circuit. It is known that the voltage at both ends of R1 is 3V, the voltage at both ends of R2 is 9V, and R1 + R2 = 400 Ω. Calculate the resistance values of R1 and R2, thank you
Two light bulbs marked with "220 V 100 W" are connected in series in a circuit with 220 V voltage, and the medium power in the circuit is () W. if they are connected in parallel in a circuit with 220 V voltage, the electric power of the circuit is () w 1
Two light bulbs marked with "220 V 100 W" are connected in series in the circuit with 220 V voltage, and the medium power in the circuit is (50) W. if they are connected in parallel in the circuit with 220 V voltage, the electric power of the circuit is (200) W
Let the resistance of each lamp be r, then p = u ^ 2 / R. now the two lamps are connected in series, so the actual total power is
Ptotal = u ^ 2 / (2R) = P / 2 = 100 / 2 = 50W
When they are connected in parallel, each lamp can emit light normally, so p = 2 * P = 2 * 100 = 200 watts
50W resistance is twice, power is 1 / 2
200w 100+100=200
Series R = u ^ 2 / P = u ^ 2 / 100
P=2* （U/2）^2/2*R
=25 W
Parallel = 2p = 200 W
R=U^2/P=U^2/100
P=2* （U/2）^2/2*R
=25 W
=2P= 200 W
Two light bulbs marked with "220 V 100 W" are connected in series in the circuit with 220 V voltage, and the medium power in the circuit is (50) W. if they are connected in parallel in the circuit with 220 V voltage, the electric power of the circuit is (200) W!!! One
100W 200W P = u & sup2 / R P string = 100W 1 lamp 100W both lamps work under u forehead for 200W naturally
I'm only in grade five. Maybe it's not right
The power is proportional to the square of the voltage. After two light bulbs are connected in series, each light bulb gets half the voltage, so the power is reduced to a quarter. The total power of the two bulbs is (50) W.
Two light bulbs in parallel in the circuit, the power of the circuit is the sum of them, for (200) W
Power of 1.5 square copper wire
My old house is decorated. It's hard to change the line. The original copper wire is 1.5 square meters. I'm going to use a refrigerator, a microwave oven and a range hood. Can I bear it?
The power of microwave oven is about 1000W, the power of range hood is about 200W, the power of refrigerator is about 200W, the total power is about 1400W
The current carrying capacity of 1.5 square copper wire is 14a, and the power it can bear is p = 14 * 220 = 3080w
1.5 square line can bear
There is an ammeter g with full bias current I = 10mA. When it is changed to a 3V voltmeter, 990 resistors need to be connected in series. What is the internal resistance of this ammeter G?
This question is unreasonable,
Full bias current I = 10mA, series connection of 990 ohm resistor,
The voltage drop on the 990 ohm resistor is: 10mA * 990 ohm = 9.9v,
It's over 3V. How can it be changed to a 3V voltmeter?

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