A) \[400\,\,J\]
B) \[25\,\,J\]
C) \[50\,\,J\]
D) \[200\,\,J\]
Correct Answer: A
Solution :
Key Idea: When resistances are joined in parallel, potential drop across them is same. From Joule's law \[P=\frac{{{V}^{2}}}{R}\] \[\Rightarrow \] \[R=\frac{{{V}^{2}}}{P}\] Given, \[V=220\,\,volt,\,\,P=100\,\,W\] \[\therefore \] \[R=\frac{220\times 220}{100}=484\,\,\Omega \] Resistance of each piece\[=\frac{484}{2}=242\,\,\Omega \] For equivalent resistance R of the two pieces joined in parallel, the combined resistance is \[\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}=\frac{1}{242}+\frac{1}{242}\] \[\Rightarrow \] \[R=121\,\,\Omega \] Hence, energy liberated/s is \[H=\frac{{{V}^{2}}}{R}t=\frac{220\times 220}{121}=400\,\,J\]
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