9th grade physics circuit questions, 9th grade physics circuit questions

The main thing is that you didn't give me the picture, it's not good to connect.

The wire comes out from the positive pole of the battery to connect to the switch S, and the other end of the switch leads two wires to connect the lamp L1 and the lamp L2 respectively, the other end of the lamp L1 is connected to the ammeter, and the other end of the ammeter is connected to the other end of the lamp L2, and the lead wire is pulled out from the other end of the lamp L2 to connect to the negative pole of the power supply.

With a small range? Do you want a physical drawing?

The picture has not been reviewed yet, and the person who reviewed it sat on D301, believe it or not, anyway, I believe it.

Yes a ah l open circuit voltmeter measurement power supply voltage current gauge number 0 c must be wrong then it must be wrong.

Explain when the current in the circuit is the smallest, and what are the numbers of the ammeter and voltmeter?

When S1 and S2 are disconnected, and the pointer of the ammeter points to B, the resistance in the circuit is the largest, and the current is the smallest;

At this time, the ammeter is short-circuited, and the indication of the ammeter is 0

The current in the circuit is i=6 (10+9+4+2)=each electrical appliance is connected in series) The indication of the voltmeter: u=ir=

When S1 closes and S2 disconnects, P is at the B-terminus. What are the indications of ammeter and voltmeter?

At this time, the ammeter is short-circuited, and the indication of the ammeter is 0

i=U (R+R3)=6 12=,R2 is short-circuited) The number of the voltmeter: U'=ir=

When S1 is disconnected, S2 is closed. pAt the A side, what are the numbers of the ammeter and the voltmeter?

At this time, the indication of the ammeter is i=u r1=6 9=2 3a=the indication of the voltmeter is 0

When S1 disconnects and S2 disconnects the sliding rheostat R at the B terminal (the resistance in the circuit is the largest at this time), the ammeter is zero, and the voltmeter (6 R+R1+R2+R3)*R, R is equal to 10 ohms.

The ammeter is zero, and the voltmeter (6 r+r3)*r, r is equal to 10 ohms.

Ammeter 6 9, voltmeter is zero.

Note: A voltmeter is equivalent to an open circuit, and an ammeter is equivalent to a short circuit.

Solution: 1. When S1 and S2 are both disconnected, and the sliding rheostat is diced at the B terminal, the current is the smallest. Ammeter display 0, voltmeter.

2. When S1 is closed and S2 is disconnected, P is at the B terminal, the ammeter is 0, and the voltmeter is 5V.

3. When S1 is disconnected and S2 is closed. P at the A terminal, ammeter display, voltmeter 0V.

R1=10 Voltage = 6V, Current = .

When S1 is disconnected and S2 is closed, the voltmeter measures the power supply voltage, and the ammeter is connected in series with R1 and R2, so the power supply voltage U=Closed When S2 is disconnected, the voltmeter measures the voltage at both ends of R1, and the voltage at both ends of R1 is U'=2V, so R1=2V;

When S1 and S2 are closed, R2 is short-circuited, the voltmeter measures the voltage at both ends of R1, which is equal to the power supply voltage, and the ammeter measures the current through R1, that is, the dry circuit current. So voltmeter = power supply voltage u = 6v. Because r1=10, the ammeter a=6v10=.

(1) When S1 is disconnected and S2 is closed, R1 is connected in series with R2, and the voltmeter measures the power supply voltage, which is obtained by Ohm's law R1+R2=U I=6V ohms. When S1 closes and S2 is disconnected, R1 and R2 are still connected in series. The voltmeter measures the voltage at both ends of R1, i.e. U1=2V.

So r1 = 2v ohms, r2 = 4v ohms.

2) When S1 and S2 are closed, R2 is short-circuited, so the current in the circuit at this time is I=U R1=6V 10=

According to P=U2R, when only the lamps are connected in series in the circuit, P total 1=U2 rl=100W, when the lamp and the resistor are connected in series in the circuit, P total 2=U2 rl+rx, obviously P total 2 P total 1, and the actual power of the lamp is known to be 81W, then the power consumed by the resistor must be less than 19W. (Let p total 2 = 99w, 99w - 81w = 18w 19w).

Because most of the electrical energy of the bulb is converted into heat energy, the resistance of the bulb r=v2 p=484

The rated current of the bulb is i=100 220=5 11aWhen the bulb power is 81W, 81=i'^2×r, i'= 9 22, so the current in the circuit is 9 22A.

The voltage at both ends of the bulb is u=i'r=198V, so the voltage across the unknown resistor is 220-198=22V

Therefore, for the unknown resistance U=22V, I=9 22AP=UI=9W, it must be lower than 19W and choose A

First of all, we know that the rated power of the lamp is 100W, and later there is an unknown resistor in series in the original circuit. The resistance in the whole circuit is definitely higher than before, and the voltage is the same, so the electrical power in the whole circuit is smaller. The power of the lamp is 81W, and of course the power of the resistor is less than 19W.

This method is called the holistic method, which is very convenient and tried and tested.