Thank you for reporting, we will resolve it shortly
Q.
Two electric bulbs marked $25W-220V$ and $100W-220V$ are connected in series to a $440V$ supply. Which of the bulbs will fuse?
NTA AbhyasNTA Abhyas 2020
Solution:
Here we will use the formula of power $P=\frac{V^{2}}{R}$ . So resistance $R=\frac{V^{2}}{P}$ . Resistance of $25W$ bulb $R_{1}=\frac{\left(220\right)^{2}}{25}=1936ohm$ and resistance of $100W$ bulb $R_{2}=\frac{\left(220\right)^{2}}{100}=484ohm$ . When they are connected in series with a supply of $440V$ , the resultant current will be $I=\frac{440}{\left(1936 + 484\right)}=0.1819A$
Now the voltage drop across $25W$ bulb will be $V_{1}=IR=0.1819\times 1936=352V$ and the voltage drop across $100W$ bulb will be $V_{2}=0.1819\times 484=88V$ . Here because of the excessive voltage (much higher than rated value) across $25W$ bulb, it will fuse.