First one light bulb is connected to a battery. Then two are connected in series to(adsbygoogle = window.adsbygoogle || []).push({});

the same battery.

Here's the picture:

http://home.comcast.net/~chou55/bulb.gif [Broken]

A) When two are connected, the battery puts out

a) less current

b) more current

c) less voltage

d) the same current

B) Which arrangement puts out the most light?

a) The single bulb

b) The two bulbs

c) Both arrangements put out the same total amount of light

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A)

I put (a) because [itex]V=IR_eq[/itex],

so[itex]I_1 = \frac {V}{R} [/itex]

and [itex]I_2 = \frac {V}{2R} [/itex]

Therefore, the second circuit has less current.

However, I heard because it's a serial circuit, so current should not change.

But I don't know how to prove that..

B)

I got (a)

because

From [itex]P = I^2 R [/itex]

I got [itex]P_1 = \frac {V^2}{R} [/itex]

and [itex]P_2 = \frac {V^2}{2R} [/itex]

so the first circuit it brighter.

But if the current is the same for both circuit, wouldn't the total light for the second circuit = the first circuit?

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I'm kind of confused now....

Could anyone help me out?

Thank you very much!

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# Homework Help: DC circuit question

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