A) \[4\,\text{kcal}\]
B) \[2\,\text{kcal}\]
C) \[5\,\text{kcal}\]
D) \[3\,\text{kcal}\]
Correct Answer: B
Solution :
Key Idea: When resistors are combined in parallel potential difference between their ends is same. The heat produced in a wire carrying \[i\] and resistance \[R\] for time \[t\] is\[H={{i}^{2}}\,Rt\]. The \[4\,\Omega \] and \[6\,\Omega \] resistances are connected in series, hence equivalent resistance is \[R'=4\,\Omega +6\,\Omega =10\,\Omega \] Now the\[10\,\Omega \] and \[5\,\Omega \] resistors are connected parallel. Let current through \[10\,\Omega \] be\[i\]and through \[5\,\Omega \] be\[{{i}_{2}}\], also potential across ends is same, hence \[V={{i}_{1}}\,{{R}_{1}}={{i}_{2}}\,{{R}_{2}}\] \[\Rightarrow \] \[{{i}_{1}}\times 10={{i}_{2}}\times 5\] \[\Rightarrow \] \[{{i}_{2}}=2{{i}_{1}}\] \[{{H}_{1}}-i_{1}^{2}{{R}_{1}}\,t\] Is heat produced across \[4\,\Omega \] \[{{H}_{2}}=i_{2}^{2}\,{{R}_{2}}\,t\]Is heat produced across\[5\,\Omega \] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{i_{1}^{2}\,4}{{{\left( 2{{i}_{2}} \right)}^{2}}\times 5}=\frac{1}{5}\] Given, \[{{H}_{2}}=10\,kcal\] \[{{H}_{1}}=\frac{{{H}_{2}}}{5}=\frac{10}{5}=2\,kcal\]You need to login to perform this action.
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