When a light bulb is connected in series with a 40 ohm resistor and connected to a 6V power supply, it can light normally. This is a small light bulb with a power of 0.2W. What is the specification of the small light bulb? It has to be calculated that --

When a light bulb is connected in series with a 40 ohm resistor and connected to a 6V power supply, it can light normally. This is a small light bulb with a power of 0.2W. What is the specification of the small light bulb? It has to be calculated that --

Suppose: the voltage of the small bulb is u, then I = 0.2/u = (6-u) / 408 = - u ^ 2 + 6U, that is: u ^ 2-6u + 8 = 0 (U-4) (U-2) = 0U = 4 or 2. If u = 4V, the bulb current is 0.2/4 = 0.05A, and the bulb specification should be 4v0.05a. It seems that there is no bulb of this specification in China

Design circuit, two lights in parallel, the existing switch 2, a power supply, a number of wires, require only closed S1 when L1 is on, only closed S2 when two lights are not on, closed S1, S2 when two lights are on, A1 only measure the current in L1, please draw the circuit diagram
You can talk about it or take a picture

S1 is on the trunk, L1 and A1 are in series, and S2 is on the branch of L2

The amount of charge passing through wire a is three times that passing through wire B in one minute
A. The current passing through a is greater than that passing through B. the current passing through B is greater than that passing through C. The current in a and B wires is the same. D. I don't know their cross-sectional area and can't compare them

ABC: the amount of charge passing through wire a in 1 min is three times that passing through wire B. according to I = QT, the current passing through wire a is three times that passing through wire B, so a is correct and BC is wrong; D. the current passing through the conductor has nothing to do with the cross-sectional area, so D is wrong. So a is selected

For example: () kg = 3336 G

3.336