A) \[40{}^\circ \]
B) \[60{}^\circ \]
C) \[50{}^\circ \]
D) \[35{}^\circ \]
Correct Answer: C
Solution :
Let the temperature of junction be \[\theta .\] In equilibrium, rate of flow of heat through rod 1 is equal to sum of rate of flow of heat through rods 2 and 3, ie, \[{{\left( \frac{dQ}{dt} \right)}_{1}}={{\left( \frac{dQ}{dt} \right)}_{2}}+{{\left( \frac{dQ}{dt} \right)}_{3}}\] \[\therefore \] \[\frac{KA(\theta -20)}{l}=\frac{KA(60-\theta )}{l}+\frac{KA(70-\theta )}{l}\] \[\Rightarrow \] \[\theta -20=130-2\theta \] or \[3\theta =150\] \[\therefore \] \[\theta =50{{\,}^{o}}C\] \[\frac{{{N}_{s}}}{{{N}_{p}}}=\frac{{{I}_{p}}}{{{I}_{s}}}\] \[\therefore \] \[\frac{20}{100}=\frac{2}{{{I}_{s}}}\] \[\frac{1}{5}=\frac{2}{{{I}_{s}}}\] \[{{I}_{s}}=5\times 2=10A\]
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