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 \[i=\frac{P}{V}=\frac{90}{30}\] = 3 amp Let resistance \[x\] is put in series when bulb is connected to \[V=120\] volt supply. Then \[3=\frac{120}{10+x}\] or \[x=30\,\,\Omega \]You need to login to perform this action.
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