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

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

Beginner2023-02-26Added 3 answers

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

${R}_{eq}=R+[\frac{R\times R}{R+R}]\phantom{\rule{0ex}{0ex}}=R+\frac{R}{2}=\frac{3R}{2}\phantom{\rule{0ex}{0ex}}i=\frac{2E}{3R}$

Current through A$=\frac{2E}{3R}$

Current through B and C A$=\frac{1}{2}[\frac{2E}{3R}]\phantom{\rule{0ex}{0ex}}=\frac{E}{3R}$

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$=\frac{1}{3}(\frac{3E}{7R})=\frac{E}{7R}$

Hence the bulbs from brightest to dimmest can be arranged as

A > D=H > B=C > E=F=G

Hence, option (a) is correct.

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

${R}_{eq}=R+[\frac{R\times R}{R+R}]\phantom{\rule{0ex}{0ex}}=R+\frac{R}{2}=\frac{3R}{2}\phantom{\rule{0ex}{0ex}}i=\frac{2E}{3R}$

Current through A$=\frac{2E}{3R}$

Current through B and C A$=\frac{1}{2}[\frac{2E}{3R}]\phantom{\rule{0ex}{0ex}}=\frac{E}{3R}$

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$=\frac{1}{3}(\frac{3E}{7R})=\frac{E}{7R}$

Hence the bulbs from brightest to dimmest can be arranged as

A > D=H > B=C > E=F=G

Hence, option (a) is correct.