A) 30 kJ
B) 2D kJ
C) 15 kJ
D) 10 kJ
Correct Answer: C
Solution :
From Joule's law, the total energy produced by bulb is \[H=\frac{{{V}^{2}}}{R}t\] Also \[P=\frac{{{V}^{2}}}{R}\] Given, \[P=100\text{ }W,V=230\,\text{volt}\] \[\therefore \] \[R=\frac{{{(230)}^{2}}}{100}=529\,\Omega \] Hence, heat produced \[=\frac{{{V}^{2}}}{R}\times t\] or \[H=\frac{115\times 115}{529}\times 10\times 60\] \[=15\,kJ\]
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