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

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

hplssrmantcxox

Full Member
10+ Year Member
Joined
Jul 5, 2011
Messages
73
Reaction score
19
b64dmwP.png


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.

:shrug::wtf:

In series:
V=V1 + V2
I = I1 = I2
R= R1+R2

Help. -_-

Members don't see this ad.
 
Last edited:
Top