A circuit problem in junior high school physics, a circuit problem in junior high school physics

First find the resistance formula r u2 p 6v 3w resistance 12 ohms 12v 3w resistance 48 ohms.

Two in series with a total resistance of 60 ohms.

Current L1 on voltage 3V< Rated voltage Luminous dark.

L2 voltage 12V = rated voltage Normal luminescence.

First of all, we need to find out the allowable current in the circuit (that is, the rated current of L2 is the smallest), only the rated current of L2 reaches the natural light, and only L2 emits light normally.

6V 3W resistance 12 ohms 12V 3W resistance 48 ohms Two in series , total resistance 60 ohms. Current L1 on the voltage 3V L2 on the voltage 12V, L2 is the rated voltage, so only L2 shines normally!

He's in trouble ==

Using r=U2 p, we can calculate the two resistors: 12 ohms and 48 ohms.

Because in a series circuit, it can be calculated that the voltage they receive is 3V and 12V respectively according to the resistance ratio (I have used this method, but you don't know if it can be or not, there is this conclusion, you can ask the teacher).

The first voltage is less than the rated voltage, so it does not emit light normally.

6V 3W resistance 12 ohms 12V 3W resistance 48 ohms Two in series , total resistance 60 ohms. Current L1 on the voltage 3V L2 on the voltage 12V, L2 is the rated voltage, so only L2 shines normally!

Amitabha!! I'll help you look at it, and then you have to understand and then copy it...

R1 is 2 ohms, p at a, v1 = 4 v, v2 = 10 v ".

At this time, R1 is connected to the power supply in series with R2. (The sliding rheostat is short-circuited and does not work) V1 is directly connected to both ends of R1, the reading of V1 is the voltage of R1 U1=4Vv2 is directly connected to both ends of R2, and the reading of V2 is the voltage of R2 U2=10V series circuit, resistance ratio = voltage divider ratio.

r2：r1=u2：u1

r2=r1u2 u1=2*10 4=5 ohms.

The latter conditions can be dispensed with.

The difficulty of this problem is that you must be very good at drawing equivalent circuit diagrams.

Note that when drawing an equivalence diagram, the voltmeter can be treated as an unenergized component or as if it does not exist. The two points connected by the wire can be seen as one point.

The explanation on the ammeter upstairs is very clear.

To give you a brief description of the voltmeter, when the sliding blade slides to the n end, the voltmeter has become a short circuit state, and the voltage cannot be measured, when the sliding blade reaches the n end, the current directly passes through the sliding vane without going to the voltmeter.

We know that the voltmeter itself has a large resistance, and the current will find a route with a small resistance, and when the slider reaches the n terminal, the sliding resistance resistance is 0 and can be regarded as a wire.

In another way, the voltmeter measures the voltage at both ends of the sliding resistance, the power supply voltage U is constant, the total resistance becomes smaller, the current becomes larger, and the sliding resistance decreases to obtain a reduced voltage.

When the slide moves towards n, the resistance of the rheostat connected to the circuit decreases, so the total resistance of the circuit decreases.

Therefore, the number of current representations becomes larger.

The resistance of the rheostat to the circuit decreases, and the total voltage of the circuit does not change, so the partial voltage obtained by the rheostat decreases.

Therefore, the number of voltage representations becomes smaller.

When sliding to the left, the sliding rheostat resistance becomes smaller, and the resistance in the circuit becomes smaller, while the total voltage does not change. Therefore, the current indicates that the number becomes larger, and the voltage indicates that the number becomes smaller.

(l) If the circuit is working normally, the ratio of the voltmeter to the ammeter should be larger (optionally "unchanged", "larger" or "smaller") according to the above operation.

2) If there is only one fault in the circuit, please judge that the fault must be R short circuit (optional "L short circuit", "L open circuit", "R short circuit" or "R open circuit").

3) If your judgment is correct, the lamp l glows and is brighter (optionally "does not glow", "glows and brighter" or "glows but dim"). The voltmeter indicates that the indication is equal to 0 (optionally "equal to 0", "equal to 3 volts", or "0 3 volts").

The ratio of the voltmeter and the ammeter is actually the resistance value of r, and the resistance value of the right shift is larger, and the fault is smaller, so r is short-circuited, the resistance is 0, and the voltage at both ends is also 0, so the voltage and current at both ends of the lamp are increased, and the brightness is increased.

(1) The ratio of the voltmeter to the ammeter should be larger

Because p slides to the right, r increases, the voltmeter reading also increases, and the total resistance increases, the current should decrease, so the ratio of the voltmeter to the ammeter should become larger.

2) The fault must be a short circuit.

r is short-circuited, the voltage decreases to 0, while the total resistance decreases and the current increases in a series circuit. In other cases, the ratio of the voltmeter to the ammeter will not be reduced.

3) The lamp l is luminous and bright, and the indication of the voltmeter is equal to 0.

Due to the short circuit of R, the voltage is reduced to 0, so that the current increases and the power of the lamp is greater and brighter.

Constant r short circuit.

Brighter glow is equal to 0