電學~電路MC 2條(~需要解釋~
2011-12-25 2:56 am
兩個額定值分別是[ 6V，5W ] 和[ 6V ，10W ] 的燈泡A 和B 串聯至6V的電池。以下項是正確的? ( 1) A的電阻值比B大( 2)通過A的電流比B少( 3)A比B暗 A 只有 ( 1)B 只有 ( 3)C 只有 ( 1)和( 2)D 只有 ( 2)和( 3) 把20個燈泡串聯至220 V電源，每個燈泡的額定值為[ 11V，4W ] 。其中一個燈泡燒了，並用新的燈泡取而代之。新的燈泡同樣標籤為4W，但新的燈泡卻較其他的暗淡。那項是正確? ( 1)新燈泡的電阻較其他燈泡的電阻少。( 2)新橙泡兩端的電壓較每個其他燈泡兩端的電壓少( 3)新燈泡的額定電壓低於220V A 只有( 1)和( 2)B 只有( 1)和( 3)C 只有( 2)和( 3)D 只有( 1)，( 2)，( 3)
回答 (2)
2011-12-25 3:55 am
✔ 最佳答案
1. Option AResistance of A = 6^/5 ohms = 7.2 ohms
Resistance of B = 6^2/10 ohms = 3.6 ohms
Hence, statement (1) is right
Statement (2) is wrong because the two bulbs are connected in series (串聯). Current in series circuit is the same.
Statement (3) is wrong
Power = current^2 xresistance
Since current is the same, bulb A has higher resistance than bulb B. Power in A is higher than that in B. Thus bulb A is brighter.
2. Option D
Statement (1) is right.
The bulbs are in series. Current through the bulbs is the same.
Since power = current^2 x resistance
i.e. resistance = power/current^2
Less bright means lower power, because current is the same, hence lower resistance for the new bulb.
Statement (2) is right.
voltage = current x resistance
current is the same, lower resistance means lower voltage across the new bulb.
Statement (3) is right.
Rated power of new bulb = 4 w
4 = V^2/r, where V is the rated voltage and r is the resistance of the new bulb
For other bulbs, 4 = 11^2/R, R is the resistance for each of the other bulbs
hence, V^2/r = 11^2/R
V^2 = 11^2 x (r/R)
Since it is found that r < R (see statement 1)
hence, V < 11 volts
2011-12-25 3:35 am
D 只有 (1)
串連電路電流相同，A電阻較大，分壓較多，應該較光A 只有( 1)和( 2)
同理因為電阻較少分壓亦少，所以較暗
串連電路電流相同，A電阻較大，分壓較多，應該較光A 只有( 1)和( 2)
同理因為電阻較少分壓亦少，所以較暗
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