Question

Bulb $1$ and Bulb $2$ are rated ($30\text{W}-200\text{V}$) and ($60\text{W}-200\text{V}$) respectively. They are connected with a $400\text{V}$ power supply. Find which bulb will get fused if they are connected in series.

• A. Bulb $1$ will get fused.
• B. Bulb $2$ will get fused.
• C. Both will get fused.
• D. No bulb will get fused.

Answer 1

The correct option is A Bulb $1$ will get fused.

We know that resistance of bulb will be

$R=\frac{V_{rated}^2}{P_{rated}}$

$R_1=\frac{(200{)}^2}{30}\Omega ,R_2=\frac{(200{)}^2}{60}\Omega$

Since, both bulbs are in series combination, so equivalent resistance will be

$R_{eq}=R_1+R_2=\frac{(200{)}^2}{20}\Omega$

So, current in circuit will be

$i=\frac{V}{R_{eq}}=\frac{400}{(200{)}^2/20}=\frac{1}{5}\text{A}$


So voltage across the bulbs will be :

$V_1=i{R}_1=\frac{1}{5}\times \frac{{200}^2}{30}=266.67\text{V}$,

$V_2=i{R}_2=\frac{1}{5}\times \frac{{200}^2}{60}=133.33\text{V}$

Hence, voltage across first bulb will be greater than $200\text{V}$. So, it will get fused.

Hence, option $\left(a\right)$ is correct.

Why this question? Key concept: If actual voltage across a device exceeds its voltage rating, it is likely to get damaged. Tip: In series if any one bulb gets fused then others will not glow.