A) 100 W lamp will fuse
B) 40 W lamp will fuse
C) Both lamps will fuse
D) Neither lamp will fuse
Correct Answer: D
Solution :
When current\[i\]flows, across potential V, then power \[{{v}_{i}}\,,\]and \[V=i\,E\] (Ohms law) The currents required by the two lamps for their normal brightness are \[{{i}_{1}}=\frac{{{p}_{1}}}{{{V}_{1}}}=\frac{40}{220}\] \[=\frac{2}{11}=0.18\,A\] \[i=\frac{V}{{{R}_{1}}+{{R}_{2}}}\] \[=\frac{40}{1694}=0.236\] \[{{i}_{2}}=\frac{{{p}_{2}}}{{{V}_{2}}}=\frac{100}{200}=0.45A\] The resistance of the filaments are \[{{R}_{1}}-\frac{V}{{{i}_{1}}}=\frac{220\times 11}{2}=121\,\Omega \]You need to login to perform this action.
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