Question 139: AC fundamentals

Question

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/
o
C) will get heated by 10
o
C. (assume no loss
of heat to the surroudnings) is close to :

Accepted Answer

f = 750 Hz, V
rms
= 20 V,
R = 100 $$\Omega $$, L = 0.1803 H,
C = 10$$\mu $$ F, S = 2 J/ÃÂ°C
|Z| = $$\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} ight)}^2}} $$
= $$\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} ight)}^2}} $$
= $$\sqrt {{R^2} + {{\left( {2\pi fL - {1 \over {2\pi fC}}} ight)}^2}} $$
= $$\sqrt {{{(100)}^2} + {{\left( {2 	imes 3.14 	imes 750 	imes 0.1803 - {1 \over {2 	imes 3.14 	imes 750 	imes {{10}^{ - 5}}}}} ight)}^2}} $$
= 834 $$\Omega $$
In AC, power (P) = i
rms
V
rms
cos $$\phi $$
and i
rms
= $${{{V_{rms}}} \over {\left| Z ight|}}$$
Power factor (cos $$\phi $$) = $${R \over {\left| Z ight|}}$$
$$ 	herefore $$ P = $${{{V_{rms}}} \over {\left| Z ight|}}.{V_{rms}}.{R \over {\left| Z ight|}}$$
= $${\left( {{{{V_{rms}}} \over {\left| Z ight|}}} ight)^2}R$$
= $${\left( {{{20} \over {834}}} ight)^2} 	imes 100$$
= 0.0575 J/S
Also, H = Pt = S$$\Delta $$$$	heta $$
$$ \Rightarrow $$ t = $${{2\left( {10} ight)} \over {0.0575}}$$ = 348 sec

Answer

348 s

Answer

418 s

Answer

245 s

Answer

365 s

AC fundamentals

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/ o C) will get heated by 10 o C. (assume no loss of heat to the surroudnings) is close to :

Choose the correct answer: