A) 20 \[\Omega \]
B) 10 \[\Omega \]
C) 40 \[\Omega \]
D) 30 \[\Omega \]
Correct Answer: D
Solution :
Here, resistance of bulb \[{{R}_{1}}=\frac{{{V}^{2}}}{P}=\frac{30\times 30}{90}=10\Omega \] current in the bulb \[i=\frac{P}{V}=\frac{90}{30}=3amp\] Let resistance x is put in series when bulb is connected to Y = 120 volt supply. Then \[3=\frac{120}{10+x}\] or \[x=30\Omega \]You need to login to perform this action.
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