Source Conversion of ac Circuits

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Source conversion is a method in AC circuits that simplifies complicated circuits by changing voltage sources to current sources or vice versa.
When applying the methods to be discussed, it may be necessary to convert a current source to a voltage source, or a voltage source to a current source. This source conversion can be accomplished in much the same manner as for source conversion in dc circuits, except now we shall be dealing with phasors and impedances instead of just real numbers and resistors.
Source conversion of ac circuits
Fig. 1: Source conversion.

Conversions of Independent Sources

In general, the format for converting one type of independent source to another is as shown in [Fig. 1].
Example 1: Convert the voltage source of [Fig. 2(a)] to a current source.
Fig. 2: Example 1.
Solution:
$$ I = {E \over Z} = {100 V \angle 0^\circ \over 5 Ω \angle 53.13^\circ} = 20A \angle -53.13^\circ$$
Example 2: Convert the current source of [Fig. 3(a)] to a voltage source.
Fig. 3: Example 2.
Solution:
$$ \begin{split} Z &= {Z_C Z_L \over Z_C + Z_L} = {(X_C \angle -90^\circ)(X_L \angle 90^\circ) \over -j X_C + j X_L}\\ &= {(4Ω \angle -90^\circ)(6 Ω\angle 90^\circ) \over -j 4 Ω+ j 6 Ω}\\ &= { 24 Ω \angle 0^\circ \over 2 \angle 90^\circ}\\ &= 12 Ω \angle -90^\circ\\ E &= I Z = (10 A \angle 60^\circ)(12 Ω \angle -90^\circ) &= 120 V \angle -30^\circ \end{split} $$

Conversions of Dependent Sources

For dependent sources, the direct conversion of [Fig. 1] can be applied if the controlling variable (V or I) is not determined by a portion of the network to which the conversion is to be applied. For example, in [Figs. 4] and [5], $V$ and $I$, respectively, are controlled by an external portion of the network. Conversions of the other kind, where $V$ and $I$ are controlled by a portion of the network to be converted, will be considered in the next chapter.
Example 3: Convert the voltage source of [Fig. 4(a)] to a current source.
Fig. 4: Example 3.
Solution:
$$ I = {E \over Z} = {20 V \angle 0^\circ \over 5k Ω \angle 0^\circ} = (4 \times 10^{-3}V) A \angle 0^\circ$$
Example 4: Convert the current source of [Fig. 5(a)] to a voltage source.
Fig. 4: Example 4.
Solution:
$$ E = IZ = [(100I) A \angle 0^\circ][ 40k Ω \angle 0^\circ]\\ = (4 \times 10^{6}I) V \angle 0^\circ$$

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