Question

A source of $750Hz$, $30$ volt is connected in series with $100ohm$ resistance, $0.1803$ henry inductance and $100$ microfarad capacitance. Find the time in which the resistance (whose thermal capacity is $3{J/}^{o}C$ will show the rise of temperature by ${20}^{o}C$.

Answer 1

Given, $V_s=30Volts$ at $750Hz$ , $R=100ohms$

$L=0.1803Henry$ and $C=100\times {10}^{-6}Farads$

$\text{Thermal Capacity}=3J{/}^{0}C$ , $\text{Rise in temperature}={20}^{0}C$

Solution,

From thermal Capacity, Total energy at ${20}^{0}C$= E =$3\times 20=60\text{Joules}....\left(1\right)$

Now, $X_L=2\pi fL$ and $X_C=\frac{1}{2\pi fC}$

therefore, $X_L=2\times \pi \times 750\times 0.1803=849.64\Omega$

and $X_C=\frac{1}{2\times \pi \times 750\times 100\times {10}^{-6}}=2.12\Omega$

therefore, $Z=\sqrt{R^2+(X_L-X_C{)}^2}$ =$\sqrt{{100}^2+(849.64-2.12{)}^2}$ =$853.39\Omega$

Hence $Current\left(i\right)=\frac{V_S}{Z}=\frac{30}{853.39}=0.035Amperes$

$\text{Energy Dessipated in resistor R =}{i}^{2}Rt$

therefore,

$Energy={0.035}^2\times 100\times t=60\text{(From equation. 1)}$

$\text{Therefore, Time required (t)}=485.53seconds$