The voltage at both ends of a conductor is 3V, the current passing through it is 0.3A, and its resistance is 10 Ω. When the current passing through the conductor is 0.5A, its resistance is Ω

The voltage at both ends of a conductor is 3V, the current passing through it is 0.3A, and its resistance is 10 Ω. When the current passing through the conductor is 0.5A, its resistance is Ω

3 / 0.5 = 6
It's very simple. Voltage divided by current is resistance
A 220 V 1000 wde electric furnace, which is connected to 110 V circuit, the power is still 1000 W
The electric furnace wire is cut into two equal length sections, which are connected in parallel in the 110V circuit. Why? Explain in detail
Why cut the electric furnace wire into two equal lengths
The method is correct
Principle:
Calculate the resistance of electric furnace:
I=P/U=1000/220=4.54(A)
R = u / I = 220 / 4.45 = 48.4 (Ohm)
Cut into two equal sections
R parallel = 48.4 / 2 * 1 / 2 = 12.1 (Ohm)
P union = u union * u union / R union = 110 * 110 / 12.1
=1000（w)
When the resistance is constant, the power of the resistance is proportional to the square of the voltage at both ends
When the voltage was 220 V, it was equivalent to the voltage of 110 V for each half of the resistance wire. Now the voltage has changed to 110 v. only by keeping the voltage of 110 V for each half of the resistance wire unchanged, can the power be kept unchanged. After parallel connection, the voltage of each half of the resistance wire is still 110 v
This explanation is easy to understand
In addition, it can be deduced from physical formula
The electric rice cooker marked with "220 V 880 W" is connected to the 220 V circuit and works for 0.5 hours. The heat generated is_______ J?
Energy consumption___________ KWh? In two ways
Electric energy consumption: 880W * 0.5h = 0.44kwh first: 1kwh = kW * H = 10 ^ 3W * 3600s = 3.6 * 10 ^ 6J, heat generated is; 0.44kwh * 3.6 * 10 ^ 6J / kWh = 1.584 * 10 ^ 6J second: 1W = 1J / s, heat generated is; 880W * 1 J / s * 30 * 60s = 1.584 * 10 ^ 6J
When the voltage at both ends of the fixed value resistor increases by 3V and the current through R increases from 0.2A to 0.3A, the resistance value of R is?
（U+3）/R=0.3
U/R=0.2
The solution of the equations is r = 30 Ω
Connect a 220 V 1000 W electric furnace to a 110 V circuit, so that the consumed power is still 1000 W
Why can't you cut 1 / 4 of the wire directly
Rated current of electric furnace I1 = U1 / R
The current after intercepting 1 / 4 is I2 = U2 / (1 / 4) r = 2u1 / r = 2i1; if it exceeds the rated current, it cannot directly intercept 1 / 4
In parallel with the same furnace on it. Question: why can't we cut 1 / 4 of the wire directly
If the current and voltage of a 220v300w electric iron fluctuate within ± 10%, what range will its actual power change?
363 and 243w
The ratio of the voltages at the two ends of each resistor in a series circuit is equal to________
In a series circuit, the ratio of the voltages at both ends of each consumer is equal to the ratio of their resistances
Series circuit has the following characteristics:
1. The same current flows through each resistor of the circuit, I = I1 = I2 = I3
2. The sum of the voltage drops on each resistor is equal to the total voltage,
That is u = IR1 + IR2 + IR3 = U1 + U2 + U3
3. The total resistance is equal to the sum of the resistances, that is, the total resistance is: = R1 + R2 + R3
4. The voltage drop ratio of each resistor is equal to its resistance ratio, i.e
U1 / U2 = R1 / R2, U1 / U3 = R1 / R2 (partial pressure formula)
Parallel resistance has the following characteristics:
1. In other words, the resistance at both ends of U1 = U3 is equal
2. The total current in the circuit is equal to the sum of the resistance branch currents, i.e. = I1 + I2 + I3
3. The reciprocal of equivalent resistance of parallel circuit is equal to the sum of reciprocal of branch resistance
1 / = 1 / R1 + 1 / r2 + 1 / R3
4. The current through each branch is inversely proportional to its resistance, i.e. I1 / I2 = R2 / R1
;
After refitting a "220V 1000W" electric furnace wire into a 110V circuit, what should we do to keep the power at 1000W?
The power on the first floor is only half of that!
The second floor theory is correct!
My opinion is:
When connecting 110V, connect the heating wire to the N line (zero line) of the power supply from the middle, and connect the other two ends to the L line (live line). In this way, the power will not change
Pay attention to the two paragraphs do not touch together!
If the power supply voltage of a '200V, 300W' electric iron fluctuates in the range of 10%, what is the actual power range?
When p = UI = u ^ 2 / R resistance is constant, P is proportional to the quadratic power of U. when u = uo = 220 V, P = Po = 300 Wu, u = 1 = (1 + 10%) u = 1.1 u, P = P1 = (U1 / UO) ^ 2 * Po = 1.1 ^ 2 * Po = 1.21 * 300 = 363 W, u = U2 = (1-10%) u = 0.9 u, P = P2 = (U2 / UO) ^ 2 * Po = 0.9 * 0.9 * 300 = 243w, the actual power is 243w to 3 W
Set constant resistance inside the electric iron
P=U^2/R
R=200^2/300
U=180~220（V）
Actual power
Pmin=180^2/R=180^2*300/200^2=243(W)
Pmax=220^2/R=220^2*300/200^2=363(W)
Why is the sum of the voltages of the resistors in a series circuit equal to the total voltage
Does any expert know. Is this just the conclusion of the experiment?
I want to ask, how did this conclusion come out
"The total resistance of series circuit is equal to the sum of the resistances" is also the conclusion drawn from "the sum of the voltages of the resistances in series circuit is equal to the total voltage"
This problem has to start from the physical law. First of all, you have to admit that the cause of current is the existence of electric field, that is, the change of electric potential energy. Second, when the circuit is closed into a loop, outside the power supply, that is, in addition to the part of the power supply, from the positive pole to the negative pole of the power supply, the potential energy changes U1; from the positive pole to the negative pole of the power supply