Does the total voltage and current in parallel circuit change with the resistance

Does the total voltage and current in parallel circuit change with the resistance

If the power supply of the parallel circuit is close to the ideal power supply such as the power supply, the total voltage of the parallel circuit will not change with the size of the resistance, and the total current will change with the change of the resistance; if your power supply has internal resistance, the total voltage and total current will change with the size of your resistance~
How many kilowatt hours of current can a 25 square meter cable carry?
25 square aluminum core cable can carry about 100A current
The current of the copper core cable is about 14025 a
The aluminum core cable with 25 square meters can carry about 22KW under the condition of 220VAC and 38KW under the condition of 380VAC
For a 25 square copper core cable, it can carry about 31 kW at 220 VAC and 53 kW at 380 VAC
What kind of situation is the landlord in?
Connect the two lamps of "220 V 40 W" and "220 V 60 W" in series into the home circuit to calculate their actual power
According to P = u ^ 2 / R
R1 = U1 ^ 2 / P1 = (220V) ^ 2 / 40W = 1210 Ω
R2 = U2 ^ 2 / P2 = (220V) ^ 2 / 60W = 2420 / 3 Ω
Then calculate the current of the series circuit, and according to the
P = I ^ 2 * r to calculate the actual power of two lights respectively. (if the number is not easy to type, I'll give you some tips.)
R1 = 220 & sup2 / 40 = 1210 Ω R2 = 220 & sup2 / 60 = 806.67 Ω actual power P = 220 & sup2; / (1210 + 806.67) = 24W
Calculate the resistance first
What is the relationship between the change of total power and the change of current, voltage and resistance?
Series circuit, with bulb and sliding rheostat. Adjust the sliding rheostat to make the resistance larger and the current smaller. How does the total power consumption change?
Besides, I'm a novice. I'm sorry. I'm out of money
Look at the change of the total resistance of the circuit. After adding, the total resistance becomes larger and the applied voltage remains unchanged. P = (U * U) / R, so it becomes smaller
P = EI, because e does not change, so you can change P as you change the current.
16 square meters of aluminum cable 50 meters 5 kilowatts, how much electricity per day loss?
prerequisite:
1. The voltage is 220 v;
2. Single phase
After calculation, the resistance of 50m 16mm2 aluminum core cable is 0.09 ohm, and the normal working current of 5kW equipment is 22.7a
Under the above conditions, the calculated daily power consumption is: (22.7 * 22.7 * 0.09) / 1000 * 24 * 2 = 2.23 kwh
Therefore, 16 square aluminum cable 50 meters 5 kilowatts per day loss of 2.23 kilowatts
In this case, the actual voltage of the light bulb will be reduced to 198w at the next peak
First of all, P = U2 / R, so the bulb resistance R = U2 / P = 484 Ω, so the actual 198v P = U2 / r = 81 watts
Explore the relationship between current, voltage and resistance. [ask question] what is the relationship between the current passing through the conductor and the voltage at both ends of the conductor and the resistance of the conductor?
Through the analysis of Ohm's law formula, I = u / R under the same supply voltage, the larger the resistance in the circuit, the smaller the current, and the smaller the resistance in the circuit, the larger the current, which shows that the resistance has the function of current limiting
The current is directly proportional to the voltage and inversely proportional to the resistance, that is, the higher the voltage, the greater the current, the greater the resistance, the smaller the current, the smaller the resistance and the greater the current
According to a famous Ohm's law, the current through a conductor is equal to the voltage at both ends of the conductor divided by the resistance of the conductor.
The current through the conductor, the voltage at both ends of the conductor and the resistance of the conductor are I = u / R
Is that a big problem?
What is the area of three-phase four wire cable for 250 kW?
120 square meters, but it depends on the specific power supply length
If two lamps marked with "220 V & nbsp; & nbsp; 60W" and "220 V & nbsp; & nbsp; 25W" are connected in series to a 220 V home circuit, the total power consumed by the two lamps is ()
A. Less than 25wb, more than 25W, less than 60wc, equal to 85wd, less than 85W, more than 60W
∵ the total resistance in the series circuit is greater than any one of the partial resistance. When two bulbs are connected in series, the total resistance in the circuit is greater than any one of the two bulbs. According to P = u2r, the total power consumed by the two bulbs is less than the power of any one bulb in normal operation, that is, less than 25W. Therefore, select a
Voltage, current or resistance, affect the brightness of the bulb? How? Please use only r = V / I questions, when the resistance increases (through the rheostat)
A few questions, when the resistance decreases (through the rheostat), the current increases and the bulb lights up?
There are four light bulbs in a circuit. The one with the highest resistance is the brightest?
If it is in series, the current through each bulb is the same. The current is constant, and the resistance is proportional to the voltage. So the higher the resistance, the higher the voltage, and the brighter the bulb. Do you understand?
The brightness of the bulb is determined by the actual power of the lamp, P = UI. Four bulbs in parallel have the highest resistance and the darkest brightness. On the contrary, four bulbs in series have the highest resistance and the brightest height. In fact, the brightness is directly proportional to the actual power.
I = V / R, the total resistance in the circuit decreases, the current increases, and the bulb becomes bright.
In the series circuit, the lamp with the highest resistance is the brightest (of course, when the lamp can light up). In the series circuit, the lamp with the highest resistance is the brightest?? Why I = V / R? Constant current? Isn't the voltage constant? According to fklaw2's answer, it is the power W = I ^ 2 * r that determines the brightness of the light bulb. In the series circuit, the current I is equal, and if the power R is greater, it will be on. The first question I = V / R is the relationship between current and resistance with constant voltage
I = V / R, the total resistance in the circuit decreases, the current increases, and the bulb becomes bright.
In the series circuit, the lightest bulb with the highest resistance (of course, when the bulb can light up) question: in the series circuit, the lightest bulb with the highest resistance?? Why I = V / R? Constant current? Isn't the voltage constant?