Subtraction of alternating quantities in AC (alternating current) circuits involves applying the same principles as subtraction in DC circuits but taking into account the complex nature of AC voltages and currents. AC quantities are typically represented using phasors, which are complex numbers used to describe the magnitude and phase angle of the alternating quantity.
Here's how subtraction of alternating quantities is performed:
Convert to Phasor Form: Convert the AC quantities to phasor or complex form. A phasor is a complex number where the real part represents the magnitude of the quantity, and the imaginary part represents the phase angle. For example, a voltage
V with amplitude
V
m
and phase angle
θ would be represented as
∠
V
m
∠θ.
Negate the Second Quantity: If you're subtracting one AC quantity from another, you can directly negate the phasor of the quantity you're subtracting.
Perform Subtraction: Perform the subtraction of the phasors. This is the same as subtracting complex numbers. Subtract the real parts and the imaginary parts separately.
Convert Back to Time Domain: Convert the result of the subtraction back to the time domain. This involves finding the magnitude and phase angle of the resulting phasor and using trigonometric functions to determine the time-varying behavior.
Remember that in AC circuits, the negative sign in front of a phasor represents a phase shift of 180 degrees.
Here's an example:
Let's say you have two AC voltages:
1
=
100
∠
3
0
∘
V
1
=100∠30
∘
V
2
=
50
∠
−
4
5
∘
V
2
=50∠−45
∘
V
Convert both quantities to phasor form:
1
=
100
∠
3
0
∘
V
1
=100∠30
∘
2
=
50
∠
−
4
5
∘
V
2
=50∠−45
∘
Negate the second quantity:
−
2
=
−
50
∠
13
5
∘
−V
2
=−50∠135
∘
Perform subtraction:
result
=
1
−
(
−
2
)
V
result
=V
1
−(−V
2
)
result
=
100
∠
3
0
∘
−
(
−
50
∠
13
5
∘
)
V
result
=100∠30
∘
−(−50∠135
∘
)
result
=
100
∠
3
0
∘
+
50
∠
13
5
∘
V
result
=100∠30
∘
+50∠135
∘
Convert back to time domain:
result
=
10
0
2
+
5
0
2
∠
(
3
0
∘
+
13
5
∘
)
V
result
=
100
2
+50
2
∠(30
∘
+135
∘
)
result
=
111.8
∠
16
5
∘
V
result
=111.8∠165
∘
So, the subtraction of the two AC voltages results in
result
=
111.8
∠
16
5
∘
V
result
=111.8∠165
∘
V.
Keep in mind that AC circuit analysis often involves more complex scenarios, such as using phasor diagrams or complex algebra to solve circuit problems.