Wattless current class 12

Knowledge Increases By Sharing...

This article will discuss the Wattless current class 12, so let’s get started…

Wattless current

Wattless current: The current in a.c. the circuit is said to be wattless if the average power consumed in the circuit is zero. The average power of an a.c. the circuit is given by
$$P_{a v}=\varepsilon_{r m s} I_{r m s} \cos \phi$$

Wattless current class 12

The above figure shows the phase angle $\phi$ between $\varepsilon_{r m s}$ and $I_{r m s}$. The current $I_{r m s}$ can be resolved into two components:

(a) Component $I_{r m s} \cos \phi$ along $\varepsilon_{r m s}$. As the phase angle between $I_{r m s} \cos \phi$ and $\varepsilon_{r m s}$ is zero, therefore $$P_{a v}=\varepsilon_{r m s}\left(I_{r m s} \cos \phi\right) \cos 0=\varepsilon_{r m s} I_{r m s} \cos \phi$$
(b) Component $I_{r m s} \sin\phi$ normal to $\varepsilon_{r m s}$. As the phase angle between $I_{r m s} \sin \phi$ and $\varepsilon_{r m s}$ is $\frac{\pi}{2}$, therefore
$$
P_{a v}=\varepsilon_{r m s}\left(I_{r m s} \sin \phi\right) \cdot \cos \frac{\pi}{2}=0
$$
We call the component $I_{r m s} \sin \phi$ as the idle or wattless current because it does not consume any power in a.c. circuit. This happens in a purely inductive or capacitive circuit in which the voltage and current differ by a phase angle of $\frac{\pi}{2}$, i.e., $\phi=\pm \frac{\pi}{2}$, so that
$$
P_{a v}=\varepsilon_{r m s} I_{r m s} \cos (\pm \pi / 2)=0
$$
Thus the current in the circuit has no power. It flows sometimes along the voltage and sometimes against the voltage so that the net work done per cycle is zero. For example, when the secondary of a transformer is open, the current in the primary is almost wattless.

Read Also

Stay tuned with Laws Of Nature for more useful interesting content.

Was this article helpful?
YesNo
Knowledge Increases By Sharing...

Newsletter Updates

Enter your email address below and subscribe to our newsletter

5 1 vote
Rate this Article
Subscribe
Notify of
guest

0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x