A) 100 J
B) 200 J
C) 300 J
D) 400J
Correct Answer: D
Solution :
(d) 400 J Resistance of heater coil, \[R=\frac{{{V}^{2}}}{P}=\frac{200\times 200}{100}=400\,\Omega \] Resistance of either half part \[=200\,\Omega \] Equivalent resistance when both parts are connected in parallel, \[R'=\frac{200\times 200}{200+200}=100\,\Omega \] Energy liberated per second when combination is connected to a source of 200 V. \[=\frac{{{V}^{2}}}{R'}=\frac{200\times 200}{100}=400\,J\]
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