Question

A $750Hz,20V\left(rms\right)$ source is connected to a resistance of $100\Omega$, an inductance of $0.1803H$ and a capacitance of $10\mu F$ all in series. The time in which the resistance (heat capacity $2J/°C$) will get heated by $10°C$. (assume no loss of heat to the surroundings) is close to:

• A. $245s$
• B. $365s$
• C. $418s$
• D. $348s$

Answer 1

The correct option is D

$348s$

Step 1: Given data and draw the diagram

Frequency of the source, $f=750Hz$

Potential of the source, $V=20V\left(rms\right)$

Value of the resistance, $R=100\Omega$

Value of inductance, $L=0.1803H$

Value of capacitor, $C=10\mu F$

Heat capacity, $S=2J/C$

Increase in temperature, $∆\theta =10°C$

Step 2: Find the value of impedance

Value of impedance, $Z=\sqrt{{\left(R_L\right)}^2+{\left(X_L-X_C\right)}^2}$

$$\begin{gathered} =\sqrt{{\left(100\right)}^2+{\left(2\pi fL-1/2\pi fC\right)}^2} \\ =\sqrt{10000+{\left(2\times 3.14\times 750\times 0.1803-1/2\times 3.14\times 750\times 10\times {10}^{-6}\right)}^2} \\ =834\Omega \end{gathered}$$

Step 3: Find the value of power

Value of power, $P=V_{rms}\times I_{rms}\times \cos \phi$

$$\begin{gathered} =V_{rms}\times \left(V_{rms}/Z\right)\times \left(R/Z\right) \\ ={\left(V_{rms}\right)}^2\times \left(R/Z^2\right) \\ ={\left(20/834\right)}^2\times \left(100\right) \\ =0.0575J/S \end{gathered}$$

Step 4: Find the value of time

Power, $P=S∆\theta /t$

$$\begin{gathered} t=\raisebox{1ex}{$S∆\theta $}\!\left/ \!\raisebox{-1ex}{$P$}\right.=2\left(10\right)/0.0575 \\ t=348sec \end{gathered}$$

Hence, option D is correct.