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

f = 750 Hz, V_rms = 20 V, <br><br>R = 100 $\Omega$, L = 0.1803 H, <br><br>C = 10$\mu$ F, S = 2 J/°C <br><br>|Z| = $\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}$ <br><br>= $\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}}$ <br><br>= $\sqrt {{R^2} + {{\left( {2\pi fL - {1 \over {2\pi fC}}} \right)}^2}}$ <br><br>= $$\sqrt {{{(100)}^2} + {{\left( {2 \times 3.14 \times 750 \times 0.1803 - {1 \over {2 \times 3.14 \times 750 \times {{10}^{ - 5}}}}} \right)}^2}} $$ <br><br> = 834 $\Omega$ <br><br>In AC, power (P) = i_rmsV_rms cos $\phi$ <br><br> and i_rms = ${{{V_{rms}}} \over {\left| Z \right|}}$ <br><br>Power factor (cos $\phi$) = ${R \over {\left| Z \right|}}$ <br><br>$\therefore$ P = ${{{V_{rms}}} \over {\left| Z \right|}}.{V_{rms}}.{R \over {\left| Z \right|}}$ <br><br>= ${\left( {{{{V_{rms}}} \over {\left| Z \right|}}} \right)^2}R$ <br><br>= ${\left( {{{20} \over {834}}} \right)^2} \times 100$ <br><br>= 0.0575 J/S <br><br>Also, H = Pt = S$\Delta$$\theta$ <br><br>$\Rightarrow$ t = ${{2\left( {10} \right)} \over {0.0575}}$ = 348 sec