Put R1 and R2 in parallel in the circuit. The resistance of R1 is 5 ohm and that of R2 is 10 ohm. What is the total resistance

Put R1 and R2 in parallel in the circuit. The resistance of R1 is 5 ohm and that of R2 is 10 ohm. What is the total resistance

1/R1+1/R2=1/R
R=R1R2/(R1+R2)
50/15=3.33333333333333333
Resistance R1 = 12 ohm and resistance R2 form a parallel circuit. If the current passing through R1 is 1 / 4 of the total current, is R2 =? 4 ohm
When u is fixed, I1 / I2 = R2 / R1, how can I1 or I2 refer to that current? By the way, how can this formula be calculated? Just take an example
Yes, R2 = 4 ohm,
Suppose the total current of the circuit is I0, then I1 = (1 / 4) I0; I2 = (3 / 4) I0
Because R1 and R2 are in parallel, the shunt law of parallel circuit should be followed
That is, R1 / r2 = I2 / I1 = (3 / 4) I0 / (1 / 4) I0 = 3
So. R2 = (1 / 3) R1 = (1 / 3) × 12 = 4 (Ω)
A: slightly
I1 / I2 = R2 / R1 is the shunt law of parallel circuit. Its source is: U1 = U2, because U1 = i1r1; U2 = i2r2, so,
If i1r1 = i2r2, we can get I1 / I2 = R2 / R1 by changing the scale formula
How to define I1 or I2? If a circuit has two resistors R1 and R2 in parallel, the current passing through R1 can be defined as I1, and the current passing through R2 can be defined as I2. Similarly, the voltage at both ends of R1 is U1, and the voltage at both ends of R2 is U2. In short, it is specified for the convenience of distinction and avoiding confusion
It's simple
That is, remember the parallel connection: the voltages on the two resistors are equal. U=IxR
That is U1 = U2 = i1xr1 = i2xr2
Then write it in the form of scale: I1 / I2 = R2 / R1
The above formula is such that I1 / (I1 + I2) = 1:4 can be calculated as I1: I2 = 1:3 and then substituted into the above formula: I1: I2 = 1:3 = R2 / R1
R2 = 4 Ω
Electromotive force of power supply e = 5V, internal resistance R = 0.5 ohm; R1 = 12.5 ohm, R2 = 20 ohm, R3 = 30 ohm, calculate the current through R1; electric power of resistance R2
Series connection of R1 and R3
(1) 2 and R3 are connected in parallel, so the parallel resistance R23 = 12 Ω, so r = R + R1 + R23 = 25 Ω
Current through R1 I = E / R total = 0.2 a
(2) The voltage at both ends of R2 is U2 = I * R23 = 2.4 v
So P2 = U2 ^ 2 / r2 = 0.288w
U/I=r+r1+1/(1/r2+1/r3)
U2=U3=I*1/(1/r2+1/r3)
P2=U2*U2/r2
There is a small bulb marked "pz220v 40W", which is connected to the 220V home circuit,
Ask: if the actual voltage is 200V, what is the actual power of the bulb? What is the luminous condition of the bulb?
Specific calculation. 3Q is not as good as university. I don't understand.
Pz220v 40W bulb resistance:
R＝U×U/P＝220×220/40＝1210(Ω)
If the actual voltage is 200V, the actual power of the bulb is as follows:
P＝U×U/R＝200×200/1210≈33(W)
The light bulb is dark red and slightly dim
(1)R=U^2/P
=(220 ^ 2 / 40) Euro
= 1210 Euro
(2) A: because the actual voltage is 220 V, which is equal to the rated voltage, the power is 40 W, and the bulb lights normally.
What is the maximum power consumption of a 5 (30) a meter?
My father wants to buy an oven and make pastry at home. Now he likes a 2700W oven
1. Our electricity meter is "5 (30) a", can we use this 2700W oven? (please answer by calculation, don't "estimate")
2. According to the knowledge learned in our physics book, the value in brackets is the maximum current allowed to pass in a short time. Why is the difference between the maximum current 5A in normal operation and the maximum current 30A allowed to pass in a short time so different?
There is no problem, but your main switch, fuse, and circuit must be able to pass more than 2700W of current; 5A (30) refers to the normal continuous running current of the meter in the range of 5 ~ 30A. For 2700W of electrical current, you can divide the power (2700) by the voltage to get the current, and the oven is generally a heating wire, which belongs to pure
The rated voltage of household electrical appliances is generally 220 V, so when the oven is 2700 W, the current is 2700 / 220 A, about 13 A, so it can not be used
P=UI=5*220=1100
[emergency!] r = resistivity multiplied by L / S is the () formula of resistance, r = u / I is the () formula of resistance
Determinism
Inferential formula
The first empty definition, the second empty decision
Determinate form
The first is determinism
The second is the definition
There is a bulb with a rated power of 40W connected to the home circuit of 220V. (set the bulb resistance unchanged) 1. Calculate the resistance of the bulb when it normally lights up. 2. If the actual voltage is 200V, what is the actual power of the bulb? How does the bulb light up?
I=P/U=40/220=0.182A
R = u / I = 220 / 0.182 = 1208.8 ohm
If the voltage is 200V, then
I=U/R=200/1208.8=0.165A
P=UI=200x0.165=33.09W
The brightness will be dimmed
As the circuit shown in the figure, when sliding rheostat slide P slides to point a, the ammeter shows 0.5 A, when p slides to midpoint C, the ammeter shows 0.4 a, if the total power consumed by the power supply is 5 W. ask: (1) what is the voltage of the power supply? (2) What is the resistance value of resistance R1? (3) What is the maximum resistance of sliding rheostat?
(1) When p is at the midpoint C, R1 is connected in series with the sliding rheostat, and the power supply voltage U = P total I = 5w0.4a = 12.5V can be obtained from P total = IU. (2) when p is at point a, there is only R1 in the circuit, then R1 = ui1 = 12.5v0.5a = 25 Ω. (3) when p is at the midpoint C, the total resistance in the circuit is r total = R1 + RC = UI = 12.5v0.4a = 31.25 Ω
Given the length, radius and resistivity, the formula of resistance value R is obtained
Known length L, radius r, resistivity ρ,
Then r = ρ L / S = ρ L / (π R & sup2;)
When a pz220-100 bulb is connected to a 220 V home circuit, what is the power consumption for one hour?
A 0.1kw. H b100kw. H c3.6 * 10 fifth power J d7.92 * 10 fifth power J
Thouk you
Please know that there is an emergency
A, of course. Don't tell me you're looking for someone to help you with your summer homework
A
A
A
According to the title, the rated voltage of the bulb is equal to 220 V, and the actual voltage is also equal to 220 V, so the actual power of the bulb is equal to the rated power P = 100 W
According to w = Pt = 100W * 1H
=0.1kw*1h
=0.1kwh
Answer: a
Thank you for your mistakes in English
100W*1h*60=6000Wh=6KWh