Question

A \[750\,Hz\], \[20\,v\left( {rms} \right)\] the source is connected to the resistance of \[100\,\Omega \], an inductance of, and a capacitance of \[10\mu F\] all in series. The time in which the resistance (heat capacity \[2J\,\,per\,\deg \,\,celsius\]) will get heated by \[{10^ \circ }C\]. (assume no loss of heat to the surrounding) is close to:
A. \[245\,\,s\]
B. \[365\,\,s\]
C. \[418\,\,s\]
D. \[348\,\,s\]

Answer 1

Hint: In this question, we are going to use the formula of the impedance of the circuit and find the time in which resistance gets heated by \[{10^ \circ }C\].

Formula used:
We have been using the following formulas:
1. \[{P_{av}} = \dfrac{{V_{rms}^2}}{{{Z^2}}} \times R\]
2. \[\omega = 2\pi f\]
3. \[{P_{av}} = {V_{rms}}{I_{rms}}\cos \phi \]
4. \[Z = \sqrt {{R^2} + {X^2}} \]
5. \[t = \dfrac{{(TC) \times \Delta \theta }}{p}\]

Complete step by step solution:
w for an LCR circuit, the average power dissipated as heat is
\[{P_{av}} = \dfrac{{V_{rms}^2}}{{{Z^2}}} \times R\] where \[Z\] is the impedance of the circuit
Now the product of power and time equals the heat generated is
\[{X_L} = \omega \,L \\
\Rightarrow {X_L} = 2\pi f\,L \\
\Rightarrow {X_L} = 2\pi \times 750 \times 0.1803 \\
\Rightarrow {X_L} = 849.2\,\Omega \\ \]
And
\[{X_C} = \dfrac{1}{{\omega \,C}} \\
\Rightarrow {X_C} = \dfrac{1}{{2\pi f\,C}} \\
\Rightarrow {X_C} = \dfrac{1}{{2\pi \times 750 \times {{10}^{ - 5}}}} \\
\Rightarrow {X_C} = 21.2\,\,\,\Omega \\ \]
So,
\[X = {X_L} - {X_C} \\
\Rightarrow X = 849.2 - 21.2 \\
\Rightarrow X = 828\,\,\Omega \\ \]
And
\[Z = \sqrt {{R^2} + {X^2}} \\
\Rightarrow Z = \sqrt {{{(100)}^2} + {{(828)}^2}} \\
\Rightarrow Z = 834\,\,\Omega \\ \]
Now in the case of ac:
\[{P_{av}} = {V_{rms}}{I_{rms}}\cos \phi \\
\Rightarrow {P_{av}} = {V_{rms}} \times \dfrac{{{V_{rms}}}}{Z} \times \dfrac{R}{Z} \\
\Rightarrow {P_{av}} = {\left( {\dfrac{{{V_{rms}}}}{Z}} \right)^2} \times R \\ \]
By substituting all values, we get
\[{P_{av}} = {\left( {\dfrac{{20}}{{834}}} \right)^2} \times 100 \\
\Rightarrow {P_{av}} = 0.00575\,\,W \\ \]
Now we know that
\[U = P \times t \\
\Rightarrow U = mc\Delta \theta \\
\Rightarrow U = (TC)\Delta \theta \\ \]
So,
\[t = \dfrac{{(TC) \times \Delta \theta }}{p} \\
\Rightarrow t = \dfrac{{2 \times 10}}{{0.0575}}\,\sec \\
\therefore t = 348\,\,\sec \]
Therefore, the time in which the resistance will get heated by \[{10^ \circ }C\] is close to \[348\,\,\sec \].

Hence, option(D) is correct option

Note: The input current is equal to the output current in a series circuit, whereas the input voltage is equal to the output voltage in a parallel circuit. So, to estimate the impedance in a series circuit, we must calculate the resulting voltage, whereas in a parallel circuit, we must calculate the consequent current. The resultant voltage in a series circuit is a real number, whereas the consequent current in a parallel circuit is either a real or an imaginary number.