Question

The circuit below is made up using identical light bulbs. The light bulbs of maximum brightness of the following will be :-

• A. A
• B. C
• C. D
• D. E

Answer 1

Answer: (B)

If bulbs are identical then their resistances are equal.
Power $P = I ^{2} R$
current through $A , B$, is $I _{ B }= I _{ A }=\frac{ V _{0}}{2 R }$
If emf of battery is $V _{0}$
$P_{ A }=P_{ B }=\left(\frac{ V _{0}}{2 R }\right)^{2} R =\frac{ V _{0}^{2}}{4 R }$
Current through $C , I _{ C }=\frac{ V _{0}}{ R + R / 2}=\frac{2 V _{0}}{3 R }$
Current through $E , D I _{ E }= I _{ D }=\frac{ I _{ C }}{2}=\frac{ V _{0}}{3 R }$
$P_{C}=I_{C}^{2} R=\left(\frac{2 V_{0}}{3 R}\right)^{2} R=\frac{4 V_{0}^{2}}{9 R}$
$P_{E}=P_{D}=\left(\frac{I_{C}}{2}\right)^{2} R=\left(\frac{V_{0}}{3 R}\right)^{2} R=\frac{V_{0}^{2}}{9 R}$
Power in $C$ is maximum so maximum brightness.