Question

$\text{A, B, C}$ are identical bulbs. How does the brightness of $\text{A}$ and $\text{B}$ change when the switch S is closed ?

• A. The brightness of $\text{A}$ increases and that of $\text{B}$ decreases.
• B. The brightness of $\text{A}$ does not change and that of $\text{B}$ decreases.
• C. The brightness of both $\text{A}$ and $\text{B}$ decreases.
• D. The brightness of $\text{A}$ increases and that of $\text{B}$ does not change.

Answer 1

The correct option is A The brightness of $\text{A}$ increases and that of $\text{B}$ decreases.

Let the resistance of each bulb be $R$. When switch is open,

Current $i=\frac{E}{2R}$

Power of both the bulb $\text{A}$ and $\text{B}$ are :

$P_A=P_B=i^{2}R=\frac{E^2}{4R}$

Now when switches is closed.


Equiavlent Resiatance :

$R_{eq}=R+\frac{R\times R}{R+R}=\frac{3R}{2}$

So current, $i=\frac{E}{R_{eq}}=\frac{2E}{3R}$

Power, $P_A=i^{2}R=\frac{4E^2}{9R}$

Power $P_B={\left(\frac{i}{2}\right)}^{2}R=\frac{E^2}{9R}$

We know that power consumed is directly proportional to the brightness.

For Bulb $\text{A}$: Final Power $\left(\frac{4E^2}{9R}\right)$> Initial power $\left(\frac{E^2}{4R}\right)$

So, the brightness of bulb $\text{A}$ increases.

For Bulb $\text{B}$ : Final Power $\left(\frac{E^2}{9R}\right)$ $<$ Initial power $\left(\frac{E^2}{4R}\right)$

So, brightness of bulb $\text{B}$ will decreases.