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Q. Bulb $B_{1}\left(100 W - 250 V\right)$ and bulb $B_{2}\left(100 \, W \, - \, 200 V\right)$ are connected across $250 \, V$ . What is potential drop across $B_{2}$ ?
Question

NTA AbhyasNTA Abhyas 2022Current Electricity

Solution:

Resistance of bulb $B_{1}, \, R_{1}=\frac{\left(250\right)^{2}}{100}= \, 625 \, \Omega$
Resistance of bulb $B_{2}, \, R_{2}=\frac{\left(200\right)^{2}}{100}= \, 400 \, \Omega$
Solution
$V_{R_{2}}= \, \frac{R_{2}}{R_{1} + R_{2}}\times \, 250=\frac{400}{625 + 400}\times \, 250=98\,V$