Question 153: AC fundamentals

Question

A series AC circuit containing an inductor (20 mH), a capacitor (120 $$\mu $$F) and a resistor (60 $$\Omega $$) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is :

Accepted Answer

Energy dissipated in 60 Sec
= (P
avg
) $$ 	imes $$ 60
= V
rms
$$ 	imes $$ I
rms
$$ 	imes $$ cos$$\phi $$ $$ 	imes $$ 60
= V
rms
$$ 	imes $$ $${{{V_{rms}}} \over Z} 	imes $$ cos$$\phi $$ $$ 	imes $$ 60
X
L
= $$\omega $$L = 2$$\pi $$FL
= 2$$\pi $$(50)$$ 	imes $$ 20 $$ 	imes $$ 10
$$-$$3
= 2$$\pi $$ $$\Omega $$
X
C
= $${1 \over {\omega C}}$$
= $${1 \over {2\pi fC}}$$
= $${1 \over {2\pi \left( {50} ight) 	imes 120 	imes {{10}^{ - 6}}}}$$
= 26.52 $$\Omega $$
$$ 	herefore $$   X
C
$$-$$ X
L
= 20.24 $$ \simeq $$ 20
$$ 	herefore $$   Z = $$\sqrt {{{\left( {{X_C} - {X_L}} ight)}^2} + {R^2}} $$
= $$\sqrt {{{\left( {20} ight)}^2} + {{60}^2}} $$
= 20  $$\sqrt {10} $$ $$\Omega $$
Also cos$$\phi $$ = $${R \over Z}$$ = $${{60} \over {20\sqrt {10} }} = {3 \over {\sqrt {10} }}$$
$$ 	herefore $$   Energy dissipated in 60 sec
= $${{{{\left( {24} ight)}^2}} \over {20\sqrt {10} }} 	imes {3 \over {\sqrt {10} }} 	imes 60$$
= 5.17 $$ 	imes $$ 10
2
J

Answer

5.65 $$ 	imes $$ 10
2
J

Answer

2.26 $$ 	imes $$ 10
3
J

Answer

5.17 $$ 	imes $$ 10
2
J

Answer

3.39 $$ 	imes $$ 10
3
J

AC fundamentals

A series AC circuit containing an inductor (20 mH), a capacitor (120 $\mu F) and a resistor (60 \Omega $) is driven by an AC source of 24V/50 Hz. The energy dissipated in the circuit in 60 s is :

Choose the correct answer: