question_answer

A10V, 650Hz, source is connected to a series combination of \[R=100\,\Omega ,\] \[C=10\mu F\] and L = 0.15H. Find out the time in which the resistance will get heated by \[10{}^\circ C,\] if the thermal capacity of the resistance \[=20J/{}^{{}^\circ }C.\]

Answer:

7086.16 s Here, \[{{E}_{V}}=10\,V,\] \[\text{v}=650\,Hz,\] \[R=100\Omega ,\] \[C=10\mu F=10\times {{10}^{-6}}F,\]\[L=0.15H,\]\[\Delta \theta =10{}^\circ C,\] \[ms=20\text{J}{}^\circ \text{/C}\] As,\[{{X}_{L}}=2\pi \,\text{v}L=2\times \frac{22}{7}\times 650\times 0.15=612.61\Omega \] And      \[{{X}_{C}}=\frac{1}{2\pi \,\text{v}C}\]             \[=\frac{1}{2\times \frac{22}{7}\times 650\times 10\times {{10}^{-6}}}=24.48\,\Omega \]             \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\]             \[=\sqrt{{{(100)}^{2}}+{{(612.61-24.48)}^{2}}}\]             \[=596.57\,\Omega \]\[{{I}_{V}}=\frac{{{E}_{V}}}{Z}=\frac{10}{596.57}=0.0168\] As,        \[I_{V}^{2}\,Rt=(ms)\,\Delta \theta \] \[\therefore \]\[t=\frac{(ms)\,\Delta \theta }{I_{V}^{2}R}=\frac{20\times 10}{{{(0.0168)}^{2}}\times 100}=7086.16\,s\]