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As the circuit shown in the figure, R1 = R2 = 12 ohm, the power supply voltage is 18 V, how much is the total resistance after R1 and R2 are connected in parallel? What is the indication of the ammeter? ┌——│R1 │————A————— │ │ │ │ │ │ │——│R2 │—— │ │ │ │ │ Quick switch
Parallel circuit: 1 / r = 1 / R1 + 1 / r2 = 1 / 12 + 1 / 12 = 2 / 12 = 1 / 6, r = 6 Ω. The ammeter measures the current on R1 branch. In the parallel circuit, the voltage of each part is the same, I1 = u total / R1 = 18V / 12 Ω = 1
6 Euro
3 A
The resistance in parallel is 6 ohm and the current is 3 a
After parallel connection, R total = R1 * R2 / (R1 + R2) = 6
I=18/6=3A
What about the picture?
The resistance of the electric lamp is 7 ohm, and the normal working voltage is 14 v. now connect it with a rheostat in series and connect it to the power supply with a voltage of 22 v. in order to make the electric lamp work normally, what is the resistance of the rheostat in the circuit? If the rheostat specification is "4a, 10 ohm", what is the maximum current and the minimum current in the circuit?
Series rheostat needs 8V partial voltage, lamp rated current = 14 / 7 = 2A
Resistance = 8 / 2 = 4 ohm
Minimum current = 22 / (7 + 10) = 22 / 17A
The maximum current is 2a, or the lamp will burn
Turn the rheostat into 4 to make the lamp work normally.
The maximum current is about 3.14a and the minimum current is 1.29a.
The resistance of the small bulb is 20 ohm. In normal operation, the voltage at both ends of the bulb is 12 volts. Now the power supply voltage is 16 volts. To connect the lamp to the power supply,
How much resistance do you need to connect? It needs to be publicized and the problem-solving process
U = u total - UL = 16-12 = 4V
I=UL/RL=12/20=0.6A
R = u / I = 4 / 0.6 = 20 / 3 Ω
You need a 20 / 3 ohm resistor bulb in series to work properly
12V / 20 = 0.6A, 16V / 0.6A = 26.67 ohm. 26.67-20 = 6.67 ohm, small bulb has more room, series 5-7 ohm resistance can be. It is not accurate to measure the resistance of the bulb. It is necessary to measure the current when the bulb is on. According to the voltage divided by the current, the calculated resistance is more accurate.
U = u total - UL = 16-12 = 4V
I=UL/RL=12/20=0.6A
R = u / I = 4 / 0.6 = 20 / 3 equals 6.7
You need a 6.7 ohm resistor bulb in series to light up normally
What's the relationship between current, resistance, voltage and power
it's a long story
It's important for them to understand the formula
Never memorize by rote
The relationship between current, resistance and voltage is Ohm's law
I=U/R
There is a formula for calculating electric power, P = UI
And then there's the deformation formula
What formula should be used to summarize what situation
There are four general powers
There are four resistors
There are two currents
These are the most common
U=IR
P=I^2*R
P=U^2/R
P (real) / P (amount) = u (real) / u (amount)
I=U/R P=UI
"When both ends of a power supply are short circuited, the power consumed on the internal resistance of the power supply is 400W, then the maximum power that the power supply can supply to the external circuit is 400W."
Why is this sentence wrong
dizzy
When the power supply is short circuited, if the line loss is ignored and the power supply is the electrical appliance, then all the current in the circuit will be used to heat the internal resistance of the current in the form of heat loss. At this time, the power is calculated as w total
When there is a real consumer on the circuit, the internal resistance of the power supply also consumes the current, and the power loss of the power supply is W0
So the real power supply of this power supply is w, and the total - W0 is less than 400W
The resistance between the two ends of a sphere with resistivity ρ and radius r? It's really hard!
Don't tell me r = ρ * L / S!!
It's useless to decide the size of your contact point
The incandescent lamp in a family circuit is marked with the word "220V 25W". When the bulb works normally, how much energy is consumed in 5h
W=Pt=25x5x3600=450000J
Advantages and disadvantages of open circuit voltage and short circuit current method for measuring open circuit voltage and equivalent internal resistance of two terminal network with source
Open circuit measurement will make some components in the circuit do not work (such as amplification circuit), affect the whole circuit, and make the measurement distortion. The biggest harm of short circuit measurement is that it may burn the components
Given the resistivity, the radius of the ball is R. calculate the resistance of the hemisphere
Equivalent method
There is an energy-saving lamp connected to the home circuit of 220 V, and its current is 0.09 A. how many kilowatt hours does this lamp use for 5 hours
W = uit = 220vx0.09ax5h = 0.099kw. H why is it? But the multiplication is finished 99 ah, why is the answer 0.099 ah? The unit of W should not be j, how can kilowatt hour appear again
Because after multiplying, 99 is watt hour, and 0.099 is converted into kilowatt hour
The unit of electric power is watt. Joule is the unit of work
Kilowatt hour is the unit of electricity consumption
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