ImmonoKedo8bj

2023-02-25

Assuming all bulbs are identical, rank the brightness of the bulbs, from brightest to dimmest.
A) all have equal brightness
B) A > B=C > D=H > E=F=G
C)A > D=H > E=F=G > B=C
D)A > D=H > B=C > E=F=G

Jamarion Turner

The right answer is (D) A > D=H > B=C > E=F=G
Brightness of the bulb $\propto$ current through the bulb
Let R represent the resistance of each bulb
Regarding the branch containing bulbs A, B, and C
${R}_{eq}=R+\left[\frac{R×R}{R+R}\right]\phantom{\rule{0ex}{0ex}}=R+\frac{R}{2}=\frac{3R}{2}\phantom{\rule{0ex}{0ex}}i=\frac{2E}{3R}$
Current through A$=\frac{2E}{3R}$
Current through B and C A$=\frac{1}{2}\left[\frac{2E}{3R}\right]\phantom{\rule{0ex}{0ex}}=\frac{E}{3R}$
For the branch containing D,E,F,G and H
${R}_{eq}=R+\frac{R}{3}+R=\frac{7R}{3}\phantom{\rule{0ex}{0ex}}i=\frac{3E}{7R}$
Current through D and H$=\frac{3E}{7R}$
Current through E,F,G$=\frac{1}{3}\left(\frac{3E}{7R}\right)=\frac{E}{7R}$
Hence the bulbs from brightest to dimmest can be arranged as
A > D=H > B=C > E=F=G
Hence, option (a) is correct.

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