A) 40
B) 10
C) 20
D) 30
Correct Answer: D
Solution :
Resistance of the lamp \[{{R}_{0}}=\frac{{{V}^{2}}}{P}\] \[=\frac{{{(30)}^{2}}}{90}=10\Omega \] Current in the lamp \[i=\frac{30}{10}=3A\] When lamp is operated on a 120 V, then resistance \[R'=\frac{V'}{i}=\frac{120}{3}=40\Omega \] Thus, for proper glow, the resistance required to the put in series will be\[R=R'-{{R}_{0}}=\text{ }40-10=30\Omega \]You need to login to perform this action.
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