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 $I_B=I_A=\frac{V_0}{2R}$
If emf of battery is $V_0$
$P_A=P_B={\left(\frac{V_0}{2R}\right)}^{2}R=\frac{V_0^2}{4R}$
Current through $C,I_C=\frac{V_0}{R+R/2}=\frac{2V_0}{3R}$
Current through $E,D{I}_E=I_D=\frac{I_C}{2}=\frac{V_0}{3R}$
$P_C=I_C^{2}R={\left(\frac{2V_0}{3R}\right)}^{2}R=\frac{4V_0^2}{9R}$
$P_E=P_D={\left(\frac{I_C}{2}\right)}^{2}R={\left(\frac{V_0}{3R}\right)}^{2}R=\frac{V_0^2}{9R}$
Power in $C$ is maximum so maximum brightness.