Question

A 750 Hz, 20 V source is connected to as resistance of 100 Omega an inductance of 0.1803 H and a capacitance of 10 muF all in sereis.Calculate the time in which the resistance (thermalcapacity 2J//.^(@)C) wil get heated by 10^(@)C.

Answer 1

The time in which the resistance will heated by 10 °C is

Explanation:

Given :

Frequency $f = 750$ Hz

Resistance $R = 100$ Ω

Inductance $L = 0.1803$ H

Capacitance $C = 10^{-5}$ F

Since impedance of the circuit,

   $Z = \sqrt{R^{2} + (X_{L} - X_{C} )^{2} }$

Where $X_{L} = 2\pi f L$, $X_{C} = \frac{1}{2\pi f C }$

For finding inductive reactance,

⇒ $X _{L} =2 \times 3.14 \times 750 \times 0.1803$

    $X_{L} = 849.2$ Ω

For finding capacitive reactance

⇒ $X_{C} = \frac{1}{2 \times \pi \times 750 \times 10^{-5} }$

   $X_{C} = 21.23$ Ω

So,

     $Z = \sqrt{(100)^{2} +(849.2 - 21.23)^{2} }$

     $Z = 833.9$ Ω

From power equation in case of AC source,

  $P =I ^{2} R$

Where $I = \frac{V_{rms} }{Z}$

  $P = (\frac{V_{rms^{} } }{Z} )^{2} R$  

Where $V_{rms} = 20$ V

  $P = (\frac{20}{833.9} )^{2} \times 100$

  $P = 0.056 \frac{J}{s}$

From the equation of thermal capacity,

   $S = \frac{ P t}{\Delta \theta}$

Where $S =$ thermal capacity, $\Delta \theta =$ 10°C

For finding the time,

    $t = \frac{2 \times 10}{0.056}$

    $t = 357$ sec

Thus, the time in which the resistance will heated by 10 °C is 357 sec