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616 ⏐⏐⏐ THE BASIC ELEMENTS AND PHASORS
Converting from polar to rectangular form for subtraction yields
IT � 84.84 mA �60° � 42.42 mA � j 73.47 mA
I1 � 56.56 mA �0° � 56.56 mA � j 0
Then
I2 � IT � I1 � (42.42 mA � j 73.47 mA) � (56.56 mA � j 0)
and I2 ��14.14 mA � j 73.47 mA
Converting from rectangular to polar form, we have
I2 � 74.82 mA �100.89°
Converting from the phasor to the time domain, we have
I2 � 74.82 mA �100.89° ⇒
i2 � �2�(74.82 � 10 �3 ) sin(qt � 100.89°)
and i2 � 105.8 � 10 �3 sin(qt � 100.89°)
A plot of the three waveforms appears in Fig. 14.68. The waveforms
clearly indicate that iT � i1 � i2. 100.89°
i 2 = i T – i 1
105.8
– 2 �
i 2
60°
i (mA)
0°
i T
i 1
�
80
� 3
2
2
�
FIG. 14.68
Solution to Example 14.32.
14.13 COMPUTER ANALYSIS
PSpice
Capacitors and the ac Response The simplest of ac capacitive
circuits will now be analyzed to introduce the process of setting up an
ac source and running an ac transient simulation. The ac source of Fig.
120
2�
�