Here, we discuss about the alternating current (A.C).
Alternating Current:
Here's an illustration of A.C.
Graphically its as follows,
The graph above has the same shape as the graphs used to represent simple harmonic motion, and it can be interpreted in the same way. The electrons, in a wire carrying a.c thus move back and forth with SHM. The current varies like a sine wave and so it is described as sinusoidal .
Mathematically,
$$I=I_{0}sin(\omega t)$$
Where,
$\omega$=angular frequency of the supply
$I_{0}$=The peak value of the alternating current (The amplitude of the graph)
And
$$V=V_{0}sin(\omega t)$$
Where:
$V_{0}$ = The peak of the voltage
$\omega$ = angular frequency
Calculating Power in A.C,
Before we define power, it is important that we define Root Mean Square value. $I_{rms}$ or root mean square current is the average value that is equivalent to the value of the steady direct current which dissipates energy in a given resistance at the same rate as the alternating current.
If $I_{rms}$ value of an a.c. is the current , then the average power $<P>$ dissipated in a load of the resistance $R$ is given by,
$$<P>=I_{rms}^{2}R=\frac{V_{rms}^{2}}{R}$$
Mathematical relationship between the peak value $I_{0}$ of the alternating current and d.c current which delivers the same average electrical power is as follows,
$$I_{rms}=\frac{I_{0}}{\sqrt{2}}$$
Similarly for $V_{rms}$ ,
$$V_{rms}=\frac{V_{0}}{\sqrt{2}}$$
WORKED EXAMPLE
A sinusoidal p.d of peak value $25V$ is connected across $20\Omega$ resistor. What is the average power dissipated in the resistor.
STEP 1: Calculate the r.m.s value os the p.d, $V_{rms}=V_{0}/ \sqrt{2}$
$V_{rms}= 25/√2 $
$V_{rms}=17.7V$
STEP 2: Now calculate the average power dissipated using, $P=V^{2}/R$
P=17.72/20
P=15.6W
(Note: if we had used $V_{0}$ instead of Vrms we would’ve got P as $31.3W$, which is double the correct answer)
Alternating Current:
"An alternating current is the flow of electric charge that periodically reverses."
Here's an illustration of A.C.
Graphically its as follows,
 |
| image credit: bbc |
The graph above has the same shape as the graphs used to represent simple harmonic motion, and it can be interpreted in the same way. The electrons, in a wire carrying a.c thus move back and forth with SHM. The current varies like a sine wave and so it is described as sinusoidal .
Mathematically,
$$I=I_{0}sin(\omega t)$$
Where,
$\omega$=angular frequency of the supply
$I_{0}$=The peak value of the alternating current (The amplitude of the graph)
And
$$V=V_{0}sin(\omega t)$$
Where:
$V_{0}$ = The peak of the voltage
$\omega$ = angular frequency
Calculating Power in A.C,
Before we define power, it is important that we define Root Mean Square value. $I_{rms}$ or root mean square current is the average value that is equivalent to the value of the steady direct current which dissipates energy in a given resistance at the same rate as the alternating current.
If $I_{rms}$ value of an a.c. is the current , then the average power $<P>$ dissipated in a load of the resistance $R$ is given by,
$$<P>=I_{rms}^{2}R=\frac{V_{rms}^{2}}{R}$$
Mathematical relationship between the peak value $I_{0}$ of the alternating current and d.c current which delivers the same average electrical power is as follows,
$$I_{rms}=\frac{I_{0}}{\sqrt{2}}$$
Similarly for $V_{rms}$ ,
$$V_{rms}=\frac{V_{0}}{\sqrt{2}}$$
WORKED EXAMPLE
A sinusoidal p.d of peak value $25V$ is connected across $20\Omega$ resistor. What is the average power dissipated in the resistor.
STEP 1: Calculate the r.m.s value os the p.d, $V_{rms}=V_{0}/ \sqrt{2}$
$V_{rms}= 25/√2 $
$V_{rms}=17.7V$
STEP 2: Now calculate the average power dissipated using, $P=V^{2}/R$
P=17.72/20
P=15.6W
(Note: if we had used $V_{0}$ instead of Vrms we would’ve got P as $31.3W$, which is double the correct answer)
Note: In collaboration with Kartikay Buchar.
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