A) \[1\,\,cal/s\]
B) \[2\,\,cal/s\]
C) \[3\,\,cal/s\]
D) \[4\,\,cal/s\]
Correct Answer: C
Solution :
Given, heat produced per second in \[5\Omega \] resistor \[=10\,\,cal/s=10\times 4.2\,\,J/s\] So, \[\frac{{{V}^{2}}}{5}=10\times 4.2\] or \[V=\sqrt{10\times 4.2\times 5V}=\sqrt{42\times 5}V\] Potential difference across\[A\]and\[B=\sqrt{42\times 5}\] So, current in\[6\Omega \], \[I=\frac{\sqrt{42\times 5}}{10}ampere\] Hence, heat produced in it, \[H={{\left( \frac{\sqrt{42\times 5}}{10} \right)}^{2}}\times 6\,\,J/s\] \[=\frac{42\times 5}{100}\times \frac{6}{4.2}=3\,\,cal/s\]
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