Question 114: AC fundamentals

Question

In the given circuit the AC source has $\omega$ = 100 rad s
-1
. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?
![](https://imagex.cdn.examgoal.net/1ktembbry/0fb56fbf-4652-4540-b08e-ab88eea42bf1/573c77e0-125a-11ec-a4fb-5347f21481c0/file-1ktembbrz.jpeg)

Accepted Answer

${Z_C} = \sqrt {{{\left( {{1 \over {\omega C}}} ight)}^2} + {R^2}} $
$ = \sqrt {{{\left( {{1 \over {100 	imes 100 	imes {{10}^{ - 6}}}}} ight)}^2} + {{100}^2}} $
${Z_C} = \sqrt {{{(100)}^2} + {{(100)}^2}} $
$ = 100\sqrt 2 $
${Z_L} = \sqrt {{{(\omega L)}^2} + {R^2}} $
$\sqrt {{{(100 	imes 0.5)}^2} + {{50}^2}} $
$ = 50\sqrt 2 $
${i_C} = {{200} \over {{Z_C}}} = {{200} \over {100\sqrt 2 }} = \sqrt 2 $
${i_L} = {{200} \over {{Z_L}}} = {{200} \over {50\sqrt 2 }} = 2\sqrt 2 $
$\cos {\phi _1} = {{100} \over {10\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _1} = 45^\circ $
$\cos {\phi _2} = {{50} \over {50\sqrt 2 }} = {1 \over {\sqrt 2 }} \Rightarrow {\phi _2} = 45^\circ $
$I = \sqrt {I_C^2 + I_L^2} $
$ = \sqrt {2 + 8} $
$ = \sqrt {10} $
I = 3.16 A
![](https://imagex.cdn.examgoal.net/1ktjevoq1/de2c8684-26ea-4ca0-a282-96d15732e01c/2fa24d90-14fd-11ec-87b0-2f866bb7de71/file-1ktjevoq2.jpeg)

Answer

5.9 A

Answer

3.16 A

Answer

0.94 A

Answer

6 A

AC fundamentals

In the given circuit the AC source has \omega = 100 rad s -1 . Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?

Choose the correct answer: