# Kaplan: Diff. resistance light bulbs in series, which is brighter? Confused..

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#### hplssrmantcxox

##### Full Member
5+ Year Member
7+ Year Member The explanation sucks..
P=IV
P = (I^2)R

Since the current stays the same:
P1 = 60 W = I^2*R1
P2 = 100 W = I^2 *R2

So if R1 = 30 ohms for example and R2 = 25 ohms for example; but the current is the same throughout the circuit and each bulb, then the actual power going through the lower powered bulb is higher than 60 W?... I'm just confused by the wording of the explanation in terms of the math.  In series:
V=V1 + V2
I = I1 = I2
R= R1+R2

Help. -_-

Last edited:

#### milski

##### 1K member
5+ Year Member
Kaplan's explanation sucks. Bulbs are in series, current is the same through each of them. To avoid retyping the same thing again: http://forums.studentdoctor.net/showthread.php?t=977407

Explaining what a "60 watt bulb" actually is would be a great start to answering this.