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 b \[R=R-{{R}_{0}}=40-10=30\,\Omega \]You need to login to perform this action.
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