The total voltage and current of resistor R1 and R2 in series are 6 V and 1.2 A, respectively The total voltage and current of resistor R1 and R2 in parallel are 2.4 V and 2 a respectively. Calculate R1 and R2

The total voltage and current of resistor R1 and R2 in series are 6 V and 1.2 A, respectively The total voltage and current of resistor R1 and R2 in parallel are 2.4 V and 2 a respectively. Calculate R1 and R2

Series resistance = 6 △ 1.2 = 5 (Ohm) = R1 + R2
Parallel resistance = 2.4 △ 2 = 1.2 (Ohm) = (R1 × R2) / (R1 + R2)
Solution: R1 = 2 (Ohm)
R2 = 3 (Ohm)
How much power can 6 square copper wire bear?
Single core or multi-core?
Three phase is about 15 kW, single phase is about 8 kW, single strand is the same as multi strand
It is known that at 20 ℃, the resistances of copper conductor, aluminum conductor and iron conductor are 0.0175 ohm, 0.029 ohm and 0.10 ohm respectively, which are 1 meter long and 1 mm 2 thick and thin. The length relationship of copper conductor, aluminum conductor and iron conductor with the same thickness and resistance is ()
A. The longest copper conductor, the shortest aluminum conductor B, the longest aluminum conductor and the shortest copper conductor
C. The longest iron conductor, the shortest copper conductor D, the longest copper conductor and the shortest iron conductor
D
After the resistor R1 and R2 are connected in parallel to the circuit, the voltage applied at both ends is 24 v. if R1 is 80 ohm, the current passing through R2 is 0.2 a, then calculate R2,
There are four solutions,
Four solutions
First, because the voltage of each parallel branch is equal to the total voltage, the voltage at both ends of R2 is 24 V, because the current is 0.2 a
I=U/R R=U/I
R = 24 V / 0.2A = 120 ohm
Second, because the voltage of each parallel branch is equal to the total voltage, R1
The voltage at both ends is 24 v
I = u / r i = 24 V / 80 ohm = 0.3A the voltage ratio of parallel circuit is equal to the inverse ratio of current, so 0.3 / 0.2 = R2 / R1 R2 = 120 ohm
Because I always = I1 + I2
I1 = U1 / R1 = u total / R1 = 24 V / 80 ohm = 0.3A and I2 = 0.2A, so I total = 0.3A = 0.2A = 0.5A
Because r total = (R1 * R2) / (R1 + R2) = (80 * R2) / (80 + R2) (*) U total = 24 V I total = 0.5 A, R total = u total / I total = 24 V / 0.5 A = 48 ohm
R is always substituted by (*), and R2 = 120 ohm
How much power can your 2.5 square copper wire bear and how much wire can you use for 10 kilowatts?
The general estimation is: copper wire per square 5 ~ 8a, aluminum wire per square 3 ~ 5A. To strictly calculate the safe current carrying capacity, but also according to the use environment of the wire, such as: limit temperature, cooling conditions, laying conditions and other comprehensive factors
The reason why the resistance of copper wire used in the test is neglected is that 1. The resistance is less than a few percent of the ohm, 2. The wire is short, 3. The resistance is less than a few hundred ohm, and the wire is very thin
The resistance is less than a few percent of a ohm
Resistor R1 and R2 are connected in series in the circuit, the negative value of R1 is 20 ohm, the ammeter in the circuit shows 0.2A, and the voltage at both ends of resistor R2 is 2 v
Calculate the negative value of resistance R1 and power supply voltage
Because the current representation is 0.2A, it can be seen that the voltage of resistor R1 is 0.2 times 20, which is equal to 4 V, while the voltage at both ends of resistor R2 is 2 v. according to the voltage ratio equal to the resistance ratio in the series circuit, it can be concluded that the resistance of R2 is equal to 10 ohm, and the power supply voltage is equal to the sum of the voltages on the two resistors, which is 6 v
How much current can a square single strand of copper wire carry
According to the laying mode, environment, quality and length, it is recommended not to exceed 8 - 9A current
There are one iron conductor and one copper conductor of the same length, and the one with the largest resistance is?
iron
Iron That's why home circuits use copper wires to reduce resistance
If the cross-sectional area and temperature are the same, it should be iron
Material resistivity, cross-sectional area and ambient temperature all affect the actual resistance value. If other conditions are the same, the resistivity of iron is greater than that of copper.
If R1 = 8 ohm, the maximum current allowed to pass is 3 A; R2 = 6 ohm, the maximum current allowed to pass is 2 A, then (L) if R1 and R2 are connected in series
Then (L) if R1 and R2 are in series, the maximum voltage allowed to be applied at both ends of the circuit is________ Volt. At the same time, the ratio of R to R2 is____ (2) if R1 and R2 are in parallel, the maximum voltage allowed at both ends of the circuit is________ Volt
（1）28
（2）4：3
（3）12
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