mankwong 發問於 科學及數學其他 - 科學 · 1 十年前

# [al phy]83 mc q43

A 10 V,5 W light bulb in a set of Christmas tree lights (which consists of 20 bulbs in series) burns out and Jimmy goes to buy a replacement. When he gets back, he finds that although the new bulb is marked 5 W, the light it gives is very dim, although the other bulbs light up brightly.

Which of the following is possible reason for this?

1. the supply voltage has dropped below 200 V.

2. the current through the circuit is very much less than 0.5A.

3. the new bulb is designed to work at a voltage lower than 10 V.

4. the new bulb is designed to work at a voltage higher than 10 V.

i have no ans

and how to deduce??

R=V^2/P=100/5=20

i=P/V=5/10=0.5

what is fractional decrement method

THX

### 1 個解答

• 1 十年前
最愛解答

1)

First of all, the bulbs in Christmas tree lights are connected in series. So if the supply voltage has dropped below 200 V, then all bulbs will have its p.d. across each of them lower than usual and hence they will also light dimmer which contradicts the fact given in the question.

So (1) is incorrect.

2)

Similar to (1), if the current through the circuit is very much less than 0.5A, then according to the relation P = (I^2)R, all other bulbs will also consume a current much less than 0.5A. Hence they all will light up dimmer too.

3)

If the bulb is designed to work at a voltage higher than 10V, then its resistance will be higher than the remaining 19 bulbs which is 10^2/5 = 20 ohm.

So if the new bulb has a resistance higher than 20 ohm, it will make the current drawn from the supply less than 0.5A. However, using fractional decrement method, it can be determined that the decrease in the power of the remaining 19 bulbs is negligible but that in the new bulb is much when compared to its rated power.

So the new bulb lights dimmer while other light up approximately the same brightness as before.

So (3) is correct.

Hence the answer is E --&gt; (3) only

2007-01-06 00:26:07 補充：

A typo mistake in (3), it should be for LOWER than 10V.eg if the bulb is for 5V, its R = 5 ohms and R(total) = 385 ohms.Current drawn = 200/385 = 0.52APower(new bulb) = 1.35W, much lower than rated.(Fractional decrement in this Q is for calculation of current. It is not a maths method.)

資料來源： My Physics knowledge