The concept of electrical power in the third year of junior high school is related to the topic of e

p=ui=i^2*r=u^2/r

The assumption that you have is not at all true, because your resistance is regarded as fixed at the stage of the current study, and if the voltage is certain, the current will also be certain, that is, the relationship between the current and the voltage is a one-time function, u=ir

So to put it this way, the greater the current, the greater the p of the path.

The higher the voltage, the greater the p of the channel.

1.When the slider P is in the midpoint position, the R lamp and half of the R change are connected in series, and the bulb is normal at this time, then the voltage at both ends of the bulb is 6V

Available at this time. u total = 6 + (6 r lamp) * 20 2 When the slider p slides to the maximum resistance position of the rheostat, the r lamp and r are connected in series, which can be obtained.

u total = 4 + (4 r lamp) * 20

The two-way solution is solved.

Supply voltage U total = 12V

The resistance of the bulb r lamp = 10 ohms.

The rated power of the bulb.

p = u r light = (6v) 10 ohm =

2.The current of the circuit when the resistance of the rheostat connected to the circuit is the maximum value.

i=4 rlamp=4 10=

The heat generated by the passage of an electric current through a rheostat.

q=i *r becomes t=(oh * 180s = 576j

Correct option: d

Analysis: 1. Because the rated voltage of the lamp is 12V, the rated power is 6W, i.e. the power consumed by the lamp is only 6W at the rated voltage of 12V.

When the lamp is connected in series with a resistor, it is still connected to the 12V power supply, and the voltage distributed between the two ends of the lamp will not be 12V, so the power consumed by the lamp is less than 6W. So the P light P resistance is 7W. Therefore, option A is incorrect.

2. When the lamp and the resistor are connected in series, the total voltage at both ends is 12V, and the total resistance of the circuit is greater than the resistance of the lamp, according to P=U2 R, the total power of the circuit is less than the power of the lamp, so P total 6WTherefore, option B is incorrect.

3. According to the analysis in 2, the actual power of the lamp and the actual power of the resistance are less than 6W, and the power of the resistance is 1W, so the actual power of the lamp must be less than 5W.

The main thing is to apply this equation to the analysis: p=u 2 r so option d is correct.

Hope it helps.

Set the resistance of the lamp to be r, and the resistance of the series to be r

Obviously, r< (r+r).

The rated power of the lamp PO=UO2 R=6W

When the lamp and the resistor are connected in series, the total voltage of the lamp and the resistor u = the rated voltage of the lamp uo = 12V The total power of the lamp and the resistor p = U2 (R+R) and the actual power of the bulb p'+ power consumed by resistor = p<6w so p,<6-1=5w....C is wrong D versus D

The brightness of the lamp is determined by the actual electrical power, which in turn is determined by the actual voltage and the actual current. If the resistance of the three lamps is not affected by temperature, L2 and L3 are regarded as a whole, which is equivalent to a resistor R23 in series with L1. According to the characteristics of the resistance of the parallel circuit, it can be known:

The equivalent resistor R23 of L2 and L3 in parallel is smaller than the smaller resistor (R3), so R23 << R1, in the series circuit, it can be seen from P=I 2R, the current is the same, and the P with large resistance is large, so L1 is the brightest, and L2 is connected in parallel with L3, and the voltage at both ends is the same, from P=U2R, it can be seen that when the voltage is the same, the P with large resistance is actually smaller, so L3 is brighter than L2, so C< P> is chosen

C, because the current of L1 is the largest, the square of the current is multiplied by the resistance, and the resistance of L1 is the largest, so L1 has the largest power and the brightest lamp. Secondly, the L2L3 voltage is the same, the voltage square is divided by the resistance, and the L2 resistance is greater than L3, so the L2 power is smaller than the L3 power. Do you understand.

The reciprocal of the total resistance of the parallel circuit is equal to the sum of the reciprocals of each resistor, for example, two resistors R1 and R2 are connected in parallel, and the total resistance is R total:

1 r total 1 r1 + 1 r2

1 r total 1 6 + 1 4 10 24

So r total 24 10 euros.

The larger the string of resistors (greater than the resistance of any of the original resistors), and the smaller and smaller (less than the resistance of any of the original resistors).

Solution: According to Ohm's law r=u i.

r total = u i = 12v 5a =

Because the total resistance in the parallel circuit is smaller than any of the resistances, the total resistance r is total

Classmate, you got the previous steps right, but the last step was wrong.

The relationship between the total resistance of the parallel connection and each resistance is: 1 r total = 1 r1 + 1 r2

R total = r1 + r2 should not be used (this is the resistance relationship of series connection), this problem is parallel.

Series connection: Two resistors are connected in series, which is equivalent to the resistance becoming longer, so the resistance becomes larger, and the two resistors need to be added up.

Parallel: Two resistors are connected in parallel, which is equivalent to the resistance becoming thicker and the cross-sectional area becoming larger, so the resistance is the smallest, that is, it is smaller than either of the two resistors, 1 and 1000 are connected in parallel, and the total resistance is smaller than 1.

1.In the circuit shown in Figure 1, the power supply voltage is 16V, and although the specifications are marked on the small bulb L, it is not clear except for the word "12V". After the switch S is closed, adjust the sliding plate P of the rheostat so that the small bulb can emit light normally, and the indication of the ammeter is 0 8A at this time, assuming that the resistance of the small bulb remains unchanged, find:

1) The resistance value of the small bulb when it emits light normally.

2) Re-adjust the sliding blade p of the rheostat, when the indication of the ammeter is 0 6A, what is the electrical power of the small bulb?

Answer: 1(1) The resistance value of the small bulb when it emits light normally is 15;

2) The electric power of the small bulb is 0 54W.

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1.r=u p=40 ohms.

2.Power consumed: 180 3600 = time = 3min = 180s

p=w/t=1000w

3. u²=r*p

u = 1000w * 40 ohms = 40000

The square of the square is u = 200

Electric power consumed by electric kettle: 180 3600=

The actual electric power of the electric kettle when it is working: :x =x=the resistance of the electric kettle:

k = u 2 r = r = 220 * 220 1210 = 40 actual voltage: k = u 2 r = u 2 = k * r = > u = 1000 * 40 = 200