One middle school physics problem and one high school physics electricity problem

Hello! 2)i1=p/u=3w/6v= i2=p/u=3w/9v=1/3a

So L2 glows normally. The current in the circuit is 1 3A

R1 U2 P (6V*6V) 3W=12 Euro P1 I2*R (1 3A*1 3A)*12 Euro 4 3W 1)9V 1 3A*12 Euro 12V

A: The power supply voltage is 12V, and the actual power of the bulb that emits dim light is about.

Let's calculate the rated current first, L1 L2

So the normal working is l2, and the current in the circuit is Otherwise the bulb will burn out and then calculate the resistance of the bulb, and the problem will be solved.

L1 12 ohms L2 27 ohms.

The supply voltage of 13V is calculated by multiplying the current by the total resistance.

The actual power of a dimly lit bulb is calculated by multiplying the current by the lamp resistance.

Upstairs 9v 1 3a*12 euro 12v miscalculated.

1) Because the bulb L1 and the bulb L2 are connected in series, the current passing through them is the same, to make the bulb L1 shine normally, that is, the current passing through it is i=p u=3w 6v=; To make the bulb L2 shine normally, that is, the current passing through it is i=p u=3w 9V; If you want the bulb L1 to shine normally, then the bulb L2 will be burned out and >, so the bulb L2 should be made to emit normally, and the resistance of the bulb L1 is R=U2 P=(6V) 2 3W=12 ohms.

The actual voltage at both ends of the bulb L1 is U=R*I=12 ohms*, so the power supply voltage U is total=

2) The bulb that emits a dim light is the bulb L1, and its actual power p=u*i=

First of all, it is important to note that the middle longitudinal path including capacitors C and R3 is an open circuit, and no current passes through after the circuit is stabilized (because the capacitor hinders the current much more than the resistor!! The indication of the galvanometer g is 0, which is ironclad) Therefore, this circuit is actually equivalent to the parallel connection of branch 1 (formed by the series connection of R1 and R2) and branch 2 (formed by the series connection of R4 and R5!! For the convenience of the following, the black dot between R1 and R2 is E, and the black dot between R4 and R5 is F, note that since there is no current on R3, capacitor C measures the potential difference between E and F!!

For branch 2, the voltage on R4 can be obtained from the properties of the series circuit as 8V, so the potential of point F is 24V-8V=16V, (note here that the flow direction of the current is from the positive electrode to the negative pole), from the amount of electricity on the capacitor c and the size of the capacitor C, it can be seen that the voltage U on C is q C=6V, that is, E, the potential difference at point F is 6V, so that the potential of point E has two possibilities, one is 22V, and the other is 10V, (because the potential difference between E and F is 6).

1) If the potential at point E is 22V, then the voltage on R1 is 24V-22V=2V, and the voltage on R2 is 22V, so that R2=22 2*R1=110 ohms.

2) If the potential at point E is 10V, then the voltage on R1 is 24V-10V=14V, and the voltage on R2 is 10V, so that R2=14 10*R1=14 ohms.

So r2 = 110 ohms or 14 ohms.

I hope it helps, and if it helps, LZ must give points!!

The answer is not d, if it is d, then the closing switch B will be short-circuited, and the disconnected switch A and B are bright, and it is wrong to show the uproar. It's not B, it should be that the position of A in the D diagram is connected to the switch on the right side of the chaotic Min Yi Jinkai.

ABC is not right, D in the two lights are in series, is a voltage divider circuit, so the voltage at both ends of the lamp is not up to 220 volts, and the light must reach 220 volts to shine, so the light will not light up the mountain is forward, the indicator light because the voltage at both ends is very small, so it emits a faint light.

d, when the switch is disconnected, A and B are connected in series, and the resistance of B lamp is much greater than the resistance of A, and the voltage on A is far lower than the voltage of B when they are connected in series, so B can shine and A does not shine.

The switch is closed, only the light light emits, A and B are wrong, because the closing switch 2 lights both light up, and the brightness is the same (the parallel circuit is loaded with the same voltage at both ends of the lamp). Because the light is A, it can only be selected in the D and C options, and the closing switch A is short-circuited and does not emit light. D item closure switch B lamp is short-circuited and does not emit light, and the voltage loaded at both ends of it at this time is 220V, which is very bright; Disconnect the switch, A and B are connected in series, A and B Changhe divide the voltage, the voltage at both ends of the A lamp is lower than 220V, and the limb wheel is relatively dark.

ABCD, I don't know how to talk about the rush, and it should be a switch to control two lights, just like a three-way intersection.