COMPLEX NUMBERS

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Phasors and Circuit Analysis –

Examples

· Objective: Present examples of the use of phasors in circuits analysis.

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Example 1: A circuit has a resistance of 5 W in series with a reactance across an inductor of 3 W. Represent the impedance by a complex number, in phasor form. Recall: The symbol for any resistance is X.

In rectangular form, the impedance is written:

Z = 5 + 3j.

In this case, X_L = 3 W and X_C = 0

so X_L - X_C = 3 W.

Using calculator, the magnitude of Z is given by: 5.83, and the angle q (the phase difference) is given by: 30.96^o.

So the voltage leads the current by 30.96^o, as shown in the diagram.

Presenting Z as a complex number (in phasor form), we have:

Z = 5.83 Ð 30.96^o.

Example 2: A particular ac circuit has a resistor of 4W, a reactance across an inductor of 8 W and a reactance across a capacitor of 11W . Express the impedance of the circuit as a complex number in polar form.

In this case, we have: X_L - X_C = -3W

So Z = 4 - 3j in rectangular form.

Now to express it in phasor form:

Using calculator, we find r = 5 and q = -36.87^o.

[NOTE: We usually express the phase angle (when voltage lags the current) using a negative value.]

So Z = 5 Ð -36.87^o

Example 3:

If R = 10 W, L = 0.6 H, C = 200 mF and w = 50 rad/s, find the impedance and the phase difference between the current and the voltage.

Recall: m (micro) means 10^-6.

Capacitive Reactance:

Inductive Reactance X_L = wL = 50 * 0.6 = 30 W

X_L - X_C = 30 - 100 = -70 W

Z = 10 – j 70 in rectangular form.

Using calculator, |Z| = 70.71 W and q = -81.87^o.

So Z = 70.71 Ð -81.87^o

So the impedance is 70.71 W and the voltage lags the current by 81.87^o(this is the phase difference).

Exercises

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