Question 67: AC fundamentals

Question

In the circuit shown, $$L = 1\,\mu H,C = 1\,\mu F\,$$ and $$R = 1\,k\Omega .$$ They are connected in series with an a.c. source  
 $$V = {V_0}\sin \omega t$$ as shown. 
Which of the following options is/are correct?

Accepted Answer

Given V = V
0
sin $$\omega$$t
At resonant frequency, current and voltage will be in same phase i.e., $$\omega L = {1 \over {\omega C}}$$
$$	herefore$$ Resonant frequency $$\omega  = {1 \over {\sqrt {LC} }}$$ ...... (i)
Here, L = 1 $$\mu$$H = 10
$$-$$6
H, C = 10
$$-$$6
F
$$	herefore$$ $$\omega  = {1 \over {\sqrt {{{10}^{ - 12}}} }} = {10^6}$$ rad s
$$-$$1
So, option (a) is incorrect.
Since, $${X_C} = {1 \over {\omega C}},\,{X_L} = \omega L$$ ....... (ii)
For large $$\omega$$, X
C
$$\sim$$ 0, so circuit will behave like a inductive circuit.
So, option (d) is incorrect.
From Eqn. (i)
$$\omega  = {1 \over {\sqrt {LC} }}$$
At resonant frequency or the frequency at which the current will be in phase with the voltage is independent of R.
$$	herefore$$ option (b) is correct.
From Eqn. (ii), at $$\omega$$ $$\sim$$ 0, X
C
$$\sim$$ $$\infty$$ and X
L
= 0
$$	herefore$$ The current flowing through the circuit becomes nearly zero.
So, option (c) is correct.

Answer

The current will be in phase with the voltage if $$\omega  = {10^4}$$ $$rad.{s^{ - 1}}$$

Answer

The frequency at which the current will be in phase with the voltage is independent of $$R$$

Answer

At $$\omega  \sim 0$$ the current flowing  through the circuit becomes nearly zero

Answer

At $$\omega  >  > {10^6}rad.{s^{ - 1}},$$ the circuit behaves like a capacitor

AC fundamentals

In the circuit shown, $L = 1\,\mu H,C = 1\,\mu F\, and R = 1\,k\Omega . They are connected in series with an a.c. source V = {V_0}\sin \omega t$ as shown. Which of the following options is/are correct?

Choose the correct answer: