A) 0.6 W
B) 0.9 W
C) 1.6W
D) 1.8W
Correct Answer: C
Solution :
When\[\theta =180{}^\circ ,\]then thermal resistance of each branch is\[\frac{R}{2}\]. So, net thermal resistance is\[\frac{R}{4}\]. Further, Thermal current\[=\frac{\Delta T}{{{R}_{net}}}\] \[\Rightarrow \] \[1.2=\frac{\Delta T}{\left[ \frac{R}{4} \right]}\] \[\Rightarrow \] \[\Delta T=0.3R\] ...(i) When\[\theta =90{}^\circ ,\] then thermal resistance of one branch is\[\frac{R}{4}\]and that of other is\[\frac{3R}{4}\]and both are in parallel \[\Rightarrow \] \[{{R}_{net}}=\frac{\left[ \frac{R}{4} \right]\left[ \frac{3R}{4} \right]}{\frac{R}{4}+\frac{3R}{4}}=\frac{3R}{16}\] \[\Rightarrow \] \[I'=\frac{\Delta T}{\left[ \frac{3R}{16} \right]}\] \[\Rightarrow \] \[I'=\frac{(0.3R)16}{3R}\] \[\Rightarrow \] \[I'=1.6\,W\]You need to login to perform this action.
You will be redirected in
3 sec