A) 400 J
B) 25 J
C) 50 J
D) 200J
Correct Answer: A
Solution :
Key Idea: When resistances are joined in parallel, potential drop across them is same. From Joules 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 \[H=\frac{{{V}^{2}}}{R}t=\frac{220\times 220\times 1}{121}=400\,J\]
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