If there is only one ammeter or voltmeter (lack of measuring tools), how to use the constant resistance or resistance box to measure the resistance It's better to give the answer before today's point

If there is only one ammeter or voltmeter (lack of measuring tools), how to use the constant resistance or resistance box to measure the resistance It's better to give the answer before today's point

If there is only one voltmeter, connect the constant resistance or resistance box in series with the resistance to be measured
First, use a voltmeter to set the voltage at both ends of the resistor and record the number U1
Then use the voltmeter to measure the resistance and add the whole voltage of the resistance to be measured, and record the indication U2
So the voltage at both ends of the resistance to be measured is UX = u2-u1
Because the resistance R0 of the constant resistance is known, and the voltage U1 at both ends of the constant resistance is measured
The current of series circuit can be calculated as I = U1 / R0
Because it is in series, the current of the resistance to be measured is the same as that of the constant resistance
So we know that the voltage at both ends of the resistance to be measured is UX = u2-u1, and the current is I = U1 / R0
So the resistance to be measured is RX = UX / I = (u0-u1) * R0 / U1
If there is only one ammeter
At this time, the resistance to be measured and the constant resistance are connected in parallel
Use an ammeter to measure the current of the constant resistance and the resistance to be measured respectively, and record the current A1 and A1 of the constant resistance respectively
And the current A2 of the resistance to be measured
Therefore, the voltage at both ends of the resistance to be measured is u = A2 * R0
In parallel circuit, the voltage is equal, so the resistance to be measured is
Rx=U/A1=A2*R0/A1

When the value of X is______ 2 = 0.4; when the value of M is 0______ 2 m = m2

(1) The & nbsp & nbsp; & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp; the & nbsp & nbsp & nbsp & nbsp & nbsp; the & nbsp & nbsp & nbsp & nbsp; & & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp; & nbsp & nbsp; & nbsp & nbsp & nbsp; & nbsp & nbsp & nbsp & amp & nbsp & nbsp; the & nbsp & nbsp; the & nbsp & nbsp & nbsp; the & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp; the & nbsp & nbsp & nbsp & nbsp & nbsp; & & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp; & & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp & nbsp A: when the value of X is 0.04, 5x + 0.2 = 0.4. (2) 2m = M2 & nbsp; & nbsp; 2m-2m = m2-2m & nbsp; m2-2m = 0m × (m-2) = 0, so m = 0 or m-2 = 0, so m = 0 or M = 2. Answer: when m is 0 or 2, 2m = m2. So the answer is: 0.04, 0 or 2

In the circuit shown in the figure, the power supply voltage is u = 60V, the internal resistance is ignored, the resistance R is 2 ohm, and the internal resistance R1 of the motor is 1.6 ohm,
The indication of the voltmeter is 50V. Calculate the output power of the motor (R is on the main circuit, R1 is in parallel with the voltmeter, and the picture can't be drawn,

If motor voltage is 50U, R voltage is 60-50 = 10u, circuit current is 10 / 2 = 5a, motor input power is 5 * 50 = 250W, thermal power is 5 * 1.6 = 8W, output power is 250-8 = 242w

6 out of 7 minus x = 2 out of 7 minus 1 out of 4,

6 out of 7 minus x = 2 out of 7 minus 1 out of 4
6 out of 7 minus x = 1 out of 28
X = 6 / 7 minus 1 / 28
X = 23 / 28

Help me figure out how much electricity I need! The power consumption of the refrigerator is 0.44 kwh / 24 hours
How much electricity does it need? How many kilowatts of electricity does it need in a month!

If the refrigerator works 24 hours a day for 30 days a month, it should be 13.2 kilowatt hours. The actual situation is that the refrigerator can consume electricity for half a day, and the working refrigerator is already a rotten one. All depends on the specific operation of your refrigerator

1 & # 178; + 2 & # 178; + 3 & # 178; + 1000 & # 178; =? Please help me

According to 1 & # 178; + 2 & # 178; + 3 & # 178; + n & # 178; = n (n + 1) (2n + 1) / 6
1²+2²+3².+1000²=1000×﹙1000＋1﹚×﹙2×1000＋1﹚÷6=333833500

If a "220 V 40 W" and B "220 V 25 W" bulbs are connected in series in a 220 V home circuit, which one is brighter? Why?

R a = u & # 178 / p a = 220 & # 178 / 40 = 1210 Ω
R B = u & # 178 / P B = 220 & # 178 / 25 = 1936 Ω
R a

How to calculate 3 out of 4 * (8 out of 18-5 out of 9)

3 / 4 * (8 / 18-5 / 9)
=3 / 4 × 8 / 18-3 / 4 * 5 / 9
=1/3-5/4
=4/12-15/12
=-11/12

The resistance wire R1 and R2 are connected in parallel. If the point in R1 loses 4 times of that in R2, the electric power consumed by R1 is P1, and the electric power consumed by R2 is P2
A.P1=P2
B.P1=2P2
C.P1=3P2
D.P1=4P3

The voltage is the same in parallel
From P = u * I, 4p1 = P2

Find an intelligence problem 9 = 72 8 = 56 7 = 42 6 = 30 5 = 20 3 =? Find
Ask an intelligence question 9 = 72 8 = 56 7 = 42 6 = 30 5 = 20 3 =? Ask why and answer

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