It is difficult to connect the 220V 100W and 220V 40W A and B in series to the 440V power supply a. They all glow normally b. The nail may burn out c. It may be burnt out d. The current works as much on the light

It is difficult to connect the 220V 100W and 220V 40W A and B in series to the 440V power supply a. They all glow normally b. The nail may burn out c. It may be burnt out d. The current works as much on the light


B may be burnt out,
Because the rated current of B is 0.1818 and that of a is 0.4545
When a and B are connected in series to the 440V power supply, the actual current is 0.2597, which is far more than the rated current of B. therefore, if C is selected, B may burn out



Given that ABCD is four different rational numbers, and (a + C) (a + D) = 1, (B + C) (b = D) = 1, find the value of (a + b) (b = C)


Analysis: (a + C) (a + D) = 1, (B + C) (B + D) = 1
Construction equation (x + C) (x + D) = 1
The equation is expanded and sorted out: X & sup2; + (c + D) x + CD-1 = 0
According to Weida's theorem, a + B = - (c + D) is a + D = - (B + C)
∴(a+c)(b+c)=-(a+c)(a+d)=-1



1. The electromotive force E of a DC power supply is 10V. The internal resistance R0 is 1 ohm. When the external resistance forms a closed circuit, what is the maximum power consumed by the external resistance?


The derivation is as follows: P = [E / (R0 + R)] ^ 2R, the numerator and denominator divide by R at the same time, the numerator becomes e ^ 2, the denominator becomes R0 + R + R0 ^ 2 / R, according to the basic inequality, R + R0 ^ 2 / R ≥ 2r0, if and only if R0 = R, the external power is the largest. Therefore, the maximum power of external resistance is 25W



Finding the differential dy of the function y = TaNx + X & # 178


dy=y'dx
=(1+tan^2x+2x)dx



Air switch single phase meter wiring diagram
How to connect a single meter from the small air switch? 1. It's the live wire incoming line, 2. It's the live wire outgoing line. The incoming line comes down from the air switch. Where should the live wire from the meter go?


The electric energy meter needs to be connected with voltage and current. The hot wire coming in and out is to connect the current. The hot wire coming out is connected with the load (electrical equipment). In addition, it needs to be connected with the zero wire to make the meter collect the voltage between the hot wire and the zero wire



Find the value of the algebraic formula X & # 178; (x-1) - x (X & # 178; + x-1), where x = - 1 / 2 (negative half). Thank you for your prompt answer


x²﹙x-1﹚-x﹙x²+x-1﹚
=x³-x²-x³-x²+x
=-2x²+x
When x = - 1 / 2
simple form
=-1/2-1/2
=-1



Under the standard condition, 500 volumes of HCl are dissolved in 1 volume of water to reach saturation, and the density of the solution is pg / ml, then the mass concentration of the solution is (mol / L)
n(HCl)=500/22.4
V=(1000 + 36.5*500/22.4)/p mL
c=n/V(L)
This is the answer on the Internet. Where did 1000 come from?


You can assume that 500 l of HCl gas is dissolved in 1 l of water
Then the amount of solute = 500L / 22.4l/mol
Mass of solution = mass of solvent + mass of solute
=1000g + 500L/22.4L/mol*36.5g/mol
Volume of solution = mass / density of solution = (1000 + 500 / 22.4 * 36.5) / P (ML)
Finally, the volume of the solution is converted into liters, and the concentration of the substance can be obtained



Let "△" be a new operation rule, 2 △ 3 = 2 + 3 + 4 = 9, 5 △ 4 = 5 + 6 + 7 + 8 = 26 According to this rule: 7 △ 5=______ .


According to the meaning of the question: 7 △ 5, = 7 + 8 + 9 + 10 + 11, = 45



As shown in the figure, there is a small bulb with a rated voltage of 2.4V. The resistance of the small bulb is 8 Ω when it works normally. If we only have a 3V power supply, how large a resistance needs to be connected in series to make the small bulb work normally?


The current in the circuit I = ulrl = 2.4v8 Ω = 0.3A, the voltage at both ends of the resistor ur = u-ul = 3v-2.4v = 0.6V, so the resistance value of the series resistor R = URI = 0.6v0.3a = 2 Ω. A: a 2 Ω resistor needs to be connected in series



4 - (- 2) ^ 3-3 ^ 2 ÷ (- 1) ^ 2N-1 + 0 * (- 2) ^ 3 Calculation


4-(-2)^3-3^2÷(-1)^2n-1+0*(-2)^3
=4+8-9÷(-1)+0
=12+9
=21