Junior high school physics: Help me solve the seventh problem

The first question is that you are not about to solve it = =

When S1 and S2 are closed, then the lamp and the slip (i.e. R2) become in parallel, and the lamp emits light normally, then we can find :

U = U Forehead Light = 12V

i lamp = p lamp u =

i2=i-i-lamp=

then you can find out of Europe.

then r2 = 20 euros.

Next is the second question (I don't know if the power supply voltage can continue to be used = =) pmax pmin=uimax uimin=rmax rmin, and because when the slip becomes 0 ohms.

i‘=>

So in the circuit, rmin=12 ohms.

rmax = 20 + 5 = 25 ohms.

So from the proportional formula obtained at the beginning, it can be obtained:

pmax pmin=rmax rmin=25 20=5 4 I don't know if it's right - -I haven't touched the question of the circuit for a long time... Ask the gods to identify it.

Analysis: (1) when S1, S2 is closed, then the bulb L is connected in parallel with the rheostat R2, and the bulb can emit light normally, and R1 is short-circuited, then the power supply voltage U, the voltage of the rheostat U=U2=UL=12V, the current value of the dry circuit measured by the ammeter I=, the current flowing through the bulb IL=PL UL=6 12A=, the resistance of the bulb RL=UL IL=12=24;

The total resistance of the rheostat R2 = 2 * U2 (I-IL) = 2 * 12 (2) When S1 and S2 are disconnected, the resistor R1 is connected in series with the rheostat R2 on the circuit, when all the rheostats are connected to the circuit, the circuit has a minimum current value I1 = U (R2 + R1) = 12 (20 + 5) A=, and the minimum power consumed by the circuit PMin=U*I1=12*

When the rheostat is not connected to the circuit, and R1 is connected to the circuit, the circuit has a maximum current value IM=U R2=12 5A=>exceeding the range of the ammeter), then when the ammeter has a maximum value i2=, the power consumed by the circuit has a maximum value pmax=u*i2=12*, then the ratio of the maximum power consumed by the circuit to the minimum power is pmax pmin=

This is the "Magic Jar" experiment.

Phenomenon: The "magic jar" that rolls to the bottom rolls back to the inclined plane.

Explanation: Moving downward, the gravitational potential energy of the "magic pot" converts kinetic energy and the elastic potential energy of the rubber band. When moving to the bottom, the elastic potential energy is the largest, and it is converted into kinetic energy, driving the "magic pot" to move upward.

The iron can rolls back, because the iron block will make the rubber band wrap around, and when it rolls down, it will store elastic potential energy, and the elastic potential energy of the rubber band at the bottom of the slope will be converted into the kinetic energy of the can and let the can roll back.

When both S1 and S2 are closed, the right part of the circuit is shorted by S1 and R2 is connected in parallel with the bulb L. At this time, the normal light emitting of the bulb means that the voltage at both ends of it is UL=6, so u=ul=u2=6V.

From Figure B, we can see that when the voltage of R2 is U2=6, the current through it i2=1a, so R2=U2 I2=6 1=6 ohms.

Total power p = pl + p2 = 3 + 6x1 = 9w.

When the voltage of the bulb is UL=3V, we can see from Figure B that the current through it IL=, so its actual power PL=ULIL=.

When both S1 and S2 are disconnected, R1 is connected in series with R2, the voltmeter measures the voltage U1 of R1, the maximum value is 3V, and the current in the ammeter circuit is I=I1=I2, and the maximum value is.

When the current in the circuit is i=, the voltage of R2 is 2=I2R2=, the voltage of R1 is u1=u-u2=, and the resistance of R1 to the circuit is connected to R1=U1 i1=ohms.

When the voltage of R1 is U1*=3V, the voltage of R2 is U2*=U- 1*=6-3=3V, and the current in the circuit is i*=I1*=I2*=U2* R2=3 6=, and R1 presses the resistance of the circuit R1*=U1* I1*=3 ohms.

The range of variation in R1 is 4 6 ohms.

Question 1 R1 is short-circuited Bulb current R2 Current R2 and bulb in parallel u=12v r 2=u i=10 r=20

Question 2 When the total resistance of the external circuit is the smallest, the total power of the circuit is the largest.

When the total resistance of the external circuit is the largest, the total power of the circuit is the smallest.

pmax pmin (u2rtotal 1) (u2r total2) rtotal2r total1 =25 5=5:1

Because q=u2 r*t means that when the voltage is constant, the smallest resistance emits the most heat, and the resistance in parallel is the smallest, so D is selected for locking

According to the formula: p u r, it can be seen that when the voltage is the same, the smaller the resistance, the greater the power, and according to the formula w pt, the greater the power, the more heat is generated.

So choose the diagram with the least resistance.

When the resistors are connected in parallel, the total resistance will be smaller, so D is selected

d Because the voltage of the four cases is equal, the total power is equal to the total voltage multiplied by the total current, it can be seen that the fourth parallel, the trunk current is the largest, so the power is the largest, and so on, the current in C is the smallest, so the power is the smallest.

First of all, the topic is clear, the power supply voltage is the same, and A, B, and C directly exclude A and B.

Look at C and D. The voltage of the parallel circuit is the same at both ends, but C is connected in series, and his calorific value is to see the current, the larger the current, the larger the resistance, the larger the heating power, and the parallel connection looks at the voltage, and the branch with small resistance has a large heating power.

p=U2 r has a small shunt resistance and high power.

A If L1 is short-circuited, since L1 and L2 are connected in parallel, L2 will also be short-circuited, so A is correct B The bulb is either short-circuited or broken, so when L1 is on, L2 can only be an open circuit fault, so B is correct.

When C L1 is not lit, according to the analysis of B answer, L1 can only be an open circuit, and the two are connected in parallel, so both ends are power supply voltages of 3V, and C is correct.

When D L1 is not lit, according to the analysis of answer B, L1 can only be an open circuit, combined with the analysis of answer C, so D is wrong.

Therefore, the correct answer is d

When the voltage is constant, the current is inversely proportional to the resistance.

The resistance becomes four times the original, and the current becomes a quarter of the original.