Put the two lamps marked 220V 40W and 220V 15W in series at 220V voltage, which lamp has the actual

The two lamps marked 220V 40W and 220V 15W are connected in series at 220V voltage, it can be said with certainty that the actual power consumption of 220V 15W lamps is large, because the voltage drop is in a section of the line with large internal resistance, and the greater the resistance, the greater the voltage drop. The larger the voltage drop, the more heat it gets, and the greater the resistance to heat, the less voltage will be obtained on a 40-watt bulb, and the current that a 40-watt lamp will pass through is only the current that a 15-watt bulb will flow through.

Therefore, the 15-watt lamp actually consumes a lot of power.

The total voltage of the two lamps after series connection u = the rated voltage u of "220V, 15W", and the total resistance of the two lamps is the resistance of "220V, 15W". Whereas.

p=ui=u 2 r, so the total power of the two lamps after series connection < the rated power of "220V, 15W" and 15W choose B

To find the resistance of 2 bulbs, first find the bulb current: 40W: i p u 40 220 15W: i p u 15 220

Find the resistance of 2 bulbs: 40W: R u i 220 15W: R u i 220

Find the current of 2 bulbs in series: i u r 220 (1209 3235) find the voltage and power at both ends of the 40w bulb: (the current of the series circuit is equal everywhere) u ri 1209

P UI Find the voltage and power at both ends of the 15W bulb:

u＝ri＝3235×

p＝ui＝162×

The two lamps of A and B are connected in series at 220V voltage, the original 15W bulb is 8W, and the original 40W bulb is 3W.

The series current is the same magnitude The power is equal to the square product resistance of the current The resistance of the B lamp is large so the power is large.

Isn't this junior high school physics? B's Great!

In the series circuit, the voltage is equal to the sum of the buried current, that is to say, the voltage of the two bulbs is 110V, and the current i = power p voltage u according to the formula; That is, i=60 110=

Now that we know the current passing through each bulb, we add the same amount of current according to the voltage of the series circuit, p = voltage 220 2*, so his total power is 60w

The answer is: d15w

60W lamp Nadan bubble resistance = 220 (60W 220V) = 807 The current of this string banquet eggplant selling circuit = 220 (807 + 807) = 220 1614 =

Voltage at both ends of a 60W bulb =

The total power of the 60W bulb = 15W

30 watts. The resistance of a light bulb is 220 (60 220) = 807 ohms.

The total current of the two bulbs in series cavity is equal to 220 (807*2) = and the power is equal to the voltage multiplied by the current, which is 220*about 30 watts of Wu Hong. Choose C

Choose C30 watts.

According to the rated voltage and power deficiency, the resistance value is calculated as 220 * 220 60 = ohms.

and then substitute into the series circuit.

With 220*[220 (watts.

Definitely digging friends!! Judge Huai.

220V, 40W bulb resistance is.

r1=U2 P=220*220 40=1210 Similarly, 220V, 25W bulb resistance is.

r2=220*220/25=1936ω

If two bulbs are connected in series, the resistance of the circuit is.

r1+r2=3146ω

So the power is.

p=u^2/r=220*220/3146=

r1=220^2/40=1210

r2=220^2/25=1936

i=220 (220 2 40+220 2 25)=power consumed by lamp B: p=

So the power consumed by lamp B is greater than 9W

This calculation is based on the assumption that the lamp resistance does not change with temperature.

r=u²/p

r1=1210ω,r1=1936ω

The two lamps of A and B are connected in series to the 220V power supply.

Then the current in the circuit i=e (r1+r2)=

B lamp power p2 = i r2 =

The first empty selection of A, the second empty selection of B.

Reason: 1. The brightness of the lamp is determined by the actual power, and the resistance of the two lamps can be calculated through the calibration data.

Because p=u2 r

So r A = (220 volts) 2 40 watts = 1210 ohms.

r B = (220 volts) 2 60 watts = 2420 3 ohms.

can get r a r b.

And because p=i2r, the two lights are connected in series, and i A = i B.

So p A p B.

When connected in series, the first light is brighter.

2. When the two lights are connected in parallel, U A = U B = 220 volts.

Therefore, the actual voltage of the two lamps is equal to the calibrated voltage, that is, the actual power is equal to the calibrated power.

From the meaning of the title, p A and p B.

Therefore, when connected in parallel, the B lamp is brighter.

Is that okay? ）

The two incandescent lamps marked with 220V, 40W and 220V, 60W are connected in series in the circuit of 220V, then (A) lamps are brighter; If the two lamps A and B are connected in parallel in the 220V circuit, then (B) lamps are brighter.

r A = 1210 ohms r B = 807 ohms, A and B two incandescent lamps have equal series current, A partial voltage and high power. A light is on.

The two lights A and B are connected in parallel in a 220V circuit, the voltage is equal B has a large power, and B lights are on.

When connected in series, the 220V-40W bulb is brighter; When connected in parallel, the 220V-60W bulb is brighter.

1. When the two lamps are connected in parallel, because the rated voltage is 220V, and the power supply voltage at this time is also 220V, they all work under the rated voltage, that is, they work normally. Therefore, their actual power is equal to the rated power, so the actual power of lamp B is large.

2. If these two lights are connected in series in a 220V circuit. r A = U2 p = 48400 40 = 1210 ohms, r B = 484 ohms.

That is, r A and r B, according to the principle of voltage division of series circuits: the voltage is proportional to the resistance. It means that the voltage at both ends of R A is large, and then according to P=I 2R, when the current of the series circuit is constant, the greater the resistance, the greater the power consumed by the electrical appliance.

Therefore, the actual power of the lamp at this time is large.

Lamp A resistance 220 * 220 100 = 484, lamp B resistance 220 * 220 25 = 1936, the current after series connection 220 (484 + 1936) =, then, lamp A voltage is 484*

The lamp B voltage is 1936*

Brother will help you summarize this kind of question. Whenever you see that you are told about the rated voltage and the rated power, you first have to think that these two conditions can be used to find the resistance from p=u 2 r to r=u 2 p. Therefore, the resistance of these two bulbs is 220V110W bulb resistance R1 = 220 * 220 110 = 440 ohms 220V40W bulb resistance R2 = 220 * 220 40 = 1210 ohms, and then this kind of question is nothing more than to put such two bulbs into a certain circuit, at this time you don't care what the previous power rating of these two lamps is, just count him as two resistors. For example, in this question, he said that it is in series, then the series current of the two resistors must be the same, at this time, choose p=i 2r to find the power, because the current is the same, you only need to compare the resistance to find the ratio of his power. So p1:

p2=ri:r2=440：1210=4：11

220V110W Bulb Resistance R1 = 220 * 220 110 = 440 ohms 220V40W Bulb Resistance R2 = 220 * 220 40 = 1210 ohms 2 are connected in series to the 220V circuit, and the current is the same. p1 = i*i*r1, p2 = i*i*r2, power ratio: p1:p2=r1:r2=440:1210=1:

The ratio of power consumed is the inverse ratio of the original power, i.e., the relationship between 4 to 11.