A) \[{{t}_{1}}+{{t}_{2}}\]
B) \[{{t}_{1}}{{t}_{2}}\]
C) \[\frac{{{t}_{1}}+{{t}_{2}}}{2}\]
D) \[\frac{{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\]
Correct Answer: D
Solution :
From Joule's law, the heat produced is \[H=\frac{{{V}^{2}}{{t}_{1}}}{{{R}_{1}}}\]and \[H=\frac{{{V}^{2}}{{t}_{2}}}{{{R}_{2}}}\] \[\Rightarrow \] \[\frac{{{t}_{1}}}{{{R}_{1}}}=\frac{{{t}_{2}}}{{{R}_{2}}}\] \[\Rightarrow \] \[\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{{{t}_{2}}}{{{t}_{1}}}\] In parallel, equivalent resistance is \[R'=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] \[H=\frac{{{V}^{2}}t}{R}\] ?(ii) \[\therefore \] Now, \[\frac{{{V}^{2}}t}{R'}=\frac{{{V}^{2}}{{t}_{1}}}{{{R}_{1}}}\] \[\Rightarrow \] \[\frac{{{t}_{1}}}{{{R}_{1}}}=\frac{t\times ({{R}_{1}}+{{R}_{2}})}{{{R}_{1}}{{R}_{2}}}\] ?(iii) From Eqs. (i) and (iii), we get \[t=\frac{{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\]
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