In this section, we take up for study an RC circuit, with the resistor and the capacitor connected in parallel. The circuit is shown in Fig. 48. Let the source be a current source, as expressed by equation (268). Given the time domain for the current as, shown by equation (268), we get the phasor expression for current, as expressed by equation (269). The task is to find expressions for the impedance of the circuit, the source voltage, the current through the resistor, and the current through the capacitor.
Since the resistor and the capacitor are in parallel, the admittance seen by the source is complex, and it is the sum of admittances of the resistor, and the capacitor. The admittance of resistor is real, and it is called the conductance. This conductance, equal to the reciprocal of the resistance, is the real part of the admittance seen by the current source. The admittance seen by the current source is called as Y.
The admittance due to the capacitor is purely imaginary, and it is called the susceptance. It is the imaginary part of the admittance seen by the current source, and it is a positive value. Equation (270) expresses the admittance seen by the source.
The impedance of the circuit is the reciprocal of its admittance. The impedance of the parallel RC network is obtained as shown by equation (271).
The phasor of the source voltage can be obtained as the phasor product of source current and impedance of the RC circuit. Since impedance is the reciprocal of admittance, it is possible to obtain the source voltage, as source current over the admittance of the RC circuit. Equation (272) presents the expression for phasor of the source voltage. Equation (273) expresses the impedance in the polar form. We can use this expression of impedance to determine the source voltage.
The phasor of the source voltage can be obtained, as the phasor product of source current and impedance of the RC circuit. Both the source current and the impedance of the circuit can be expressed in the polar form, and the phasor of the source voltage is expressed by equation (274). The time domain expression for the source voltage is presented by equation (275).
We can now determine the expression for the resistor current. Even though it is easy to obtain the resistor current, as the ratio of source voltage to the resistance, we obtain the resistor current using the current division rule. Since the resistor and the capacitor are in parallel, we can use the current division rule to find the resistor current. The resistor current is proportionate to its admittance, as shown by equation (276). Get the ratio of admittance of resistor to the sum of admittances due to the resistor and the capacitor. Multiply this ratio by the source current to get the resistor current. As shown by equation (276), the resistor current is just the ratio of the source voltage over the resistance. The time domain expression for the resistor current is expressed by equation (277).
We can obtain an expression for the phasor of capacitor current using the current division rule, as shown by equation (278). It can be expressed in terms of the admittance of the capacitor, the source current and the impedance of the circuit, as shown by equation (278). Since the product of the source current, and the impedance of the circuit is the source voltage, we get the first part of equation (278). We get the middle part of equation (278), by simplifying the expression obtained in the first part.. Capacitor current can also be expressed as the product of the admittance of capacitor and the source voltage, as shown by the last part of equation (278).
The time domain expression for capacitor current is obtained as follows. Equation (279) expresses the source voltage. The phasor representing the capacitor current is obtained by multiplying the voltage phasor by the admittance of the capacitor, as shown by equation (280). We know that the current phasor leads the voltage phasor by 90o. This is because the capacitor current is j times the product of susceptance of capacitor and phasor of source voltage. It has been proven earlier that multiplying a phasor by j rotates it in anti-clockwise direction by 90o. The time domain expression of capacitor current is displayed by equation (281).
Here an explanation for the phasor diagram in Fig. 48 is provided. We get equation (281) from the circuit in Fig. 48. Both the resistor and the capacitor are connected across the source, and their voltage equals the source voltage. The phasor sum of resistor current, and the capacitor current, is the source current, as shown by equation (282). The capacitor voltage lags the capacitor current by 90o. As shown by equation (281), the capacitor voltage is the same as the resistor voltage. Hence the resistor voltage, and the resistor current, lag the capacitor current by 90o. As shown in the phasor diagram, the source current has a phase of fo, with respect to the reference phasor, which lies on the horizontal axis, to the right of origin. The phasor diagram shows how the phasor sum of resistor current, and the capacitor current can be obtained.
It is possible to obtain the equivalent the series and the parallel connected RC circuits. Given the series connected RC circuit, we can obtain the equivalent parallel connected RC circuit by equating the admittances of the two circuits to each other.
Equation (282) equates the two admittances. Equation (283) is obtained multiplying both the numerator and the denominator of the right-hand side of equation (282) by the conjugate of the denominator. By comparing the real parts, we get Rp. By comparing the imaginary parts, we get Cp. If the resistive component of the series-connected circuit is very much larger than the magnitude of the reactive component, then we get equation (285). If the resistive component of the series-connected circuit is very much smaller than the magnitude of the reactive component, then we get equation (286).
Given the parallel connected RC circuit, we can obtain the equivalent series connected RC circuit by equating the impedances of the two circuits to each other.
Equation (287) equates the two impedances. Equation (288) is obtained multiplying both the numerator and the denominator of the right-hand side of equation (287) by the conjugate of the denominator. By comparing the real parts, we get Rs. By comparing the imaginary parts, we get Cs. If the resistive component of the parallel-connected circuit is very much larger than the magnitude of the reactive component, then we get equation (290). If the resistive component of the parallel-connected circuit is very much smaller than the magnitude of the reactive component, then we get equation (291).
In this section, we take up for study an RL circuit, with the resistor and the inductor connected in parallel. The circuit is shown in Fig. 50. Let the source be a current source, as expressed by equation (292). The task is to find expressions for the impedance of the circuit, the source voltage, the current through the resistor and the current through the inductor.
Given the time domain expression for the current as, shown by equation (292), we get the phasor expression for current, as expressed by equation (293).
Since the resistor and the inductor are in parallel, the admittance seen by the source is complex, and it is the sum of admittances of the resistor and the inductor. The admittance of resistor is real, and it is called the conductance. This conductance, equal to the reciprocal of the resistance, is the real part of the admittance seen by the current source. The admittance seen by the current source is called as Y.
The admittance due to the inductor is purely imaginary, and it is called the susceptance. It is the imaginary part of the admittance seen by the current source. Equation (294) expresses the admittance seen by the source. Since the susceptance of an inductor is negative, the imaginary part of the admittance seen by the source is negative.
The impedance of the circuit is the reciprocal of its admittance. The impedance of the parallel RL network is obtained as shown by equation (295). It looks complicated, but given numerical values for resistor, inductance and the angular frequency, it gets simplified to a complex value.
The phasor of the source voltage can be obtained as the phasor product of source current and impedance of the RL circuit. Since impedance is the reciprocal of admittance, we can obtain the source voltage as source current over the admittance of the RL circuit. Equation (296) presents the expression for phasor of the source voltage. We can express the impedance of the RL circuit as shown by the first part of equation (297), where the denominator is in Cartesian form. It can be expressed in the polar form, as shown by the second part of equation (297). We can use this expression of impedance to determine the source voltage.
The phasor of the source voltage can be obtained, as the phasor product of source current and impedance of the RL circuit. Both the source current and the impedance of the circuit can be expressed in the polar form, and the phasor of the source voltage is expressed by equation (298). The time domain expression for the source voltage is presented by equation (299).
We can now determine the expression for the resistor current. Even though it is easy to obtain the resistor current as the ratio of source voltage to the resistance, we obtain the resistor current using the current division rule. Since the resistor and the inductor are in parallel, we can use the current division rule to find the resistor current. The resistor current is proportionate to its admittance, as shown by equation (300). Get the ratio of admittance of resistor to the sum of admittances due to the resistor and the inductor. Multiply this ratio by the source current to get the resistor current. The denominator of the middle part of equation (300) is the admittance of the circuit, and its reciprocal is impedance of the circuit. As shown by the last part of equation (300), the resistor current is just the ratio of the source voltage over the resistance. The time domain expression for the resistor current is expressed by equation (301).
We can obtain an expression for the phasor of inductor current using the current division rule, as shown by equation (302). The first part of equation (302) has the admittance of the inductor as the numerator and the admittance of both R and L in parallel in the denominator. The admittance of both R and L in parallel is Y, and hence we get the second part of equation (302). Since the impedance is the reciprocal of admittance, we get the third part of equation (302). The inductor current can be expressed as the ratio of the source voltage to the impedance of the inductor, as shown by the last part of equation (302).
The time domain expression for inductor current is obtained as follows. Equation (303) expresses the source voltage. The current phasor is obtained by dividing the voltage phasor by j times the reactance of the inductor, as shown by equation (304). Because of the presence of j in the denominator of equation (304), the phasor for inductor current lags the phasor for source voltage by 90o. The time domain expression of capacitor current is displayed by equation (305).
Here an explanation for the phasor diagram in Fig. 50 is provided. Both the resistor and the inductor are connected across the source, and their voltage equals the source voltage, as shown by equation (306). The phasor sum of resistor current, and the inductor current, is the source current, as shown by equation (307). The inductor voltage leads the inductor current by 90 degrees. As shown by equation (306), the inductor voltage is the same as the resistor voltage. Hence the resistor voltage and the resistor current lead the inductor current by 90o. As shown in the phasor diagram, the source current has a phase of fo, with respect to the reference phasor, which lies on the horizontal axis, to the right of origin. The phasor sum of resistor current, and the inductor current, is the source current, as shown by equation (307). The phasor diagram shows, that the source current is the phasor sum of resistor current and the inductor current.
It is possible to obtain the equivalent the series and the parallel connected RL circuits. Given the series connected RL circuit, we can obtain the equivalent parallel connected RL circuit by equating the admittances of the two circuits to each other.
Equation (308) equates the two admittances. As shown by equation (309), we get Rp by comparing the real parts. By comparing the imaginary parts, we get Lp. If the resistive component of the series-connected circuit is very much larger than the magnitude of the reactive component, then we get equation (310). If the resistive component of the series-connected circuit is very much smaller than the magnitude of the reactive component, then we get equation (311).
Given the parallel connected RL circuit, we can obtain the equivalent series connected RL circuit by equating the impedances of the two circuits to each other.
Equation (312) equates the two impedances. Equation (313) is obtained multiplying both the numerator and the denominator of the right-hand side of equation (312) by the conjugate of the denominator. As shown by equation (314), we get Rs by comparing the real parts. By comparing the imaginary parts, we get Cs. If the resistive component of the parallel-connected circuit is very much larger than the magnitude of the reactive component, then we get equation (315). If the resistive component of the parallel-connected circuit is very much smaller than the magnitude of the reactive component, then we get equation (316).
In this section, a brief discussion on LC circuits is presented. Fig. 52 shows an inductor and a capacitor connected in series. The impedance of this network is presented by equation (317) and it has only reactive component. If the net reactance is positive, the impedance has inductive reactance, as shown by equation (318). If the net reactance is negative, the impedance has capacitive reactance, as shown by equation (319).
The admittance of the series LC circuit is the reciprocal of its impedance, as expressed by equation (320). When w2 equals the reciprocal of the product of inductance and capacitance, the admittance is infinite and the impedance is zero. At this frequency, the series network consisting of the inductor and the capacitor is a short circuit, at viewed from its terminals. Equation (321) states that the admittance is infinite if w has the specified value. This condition leads to resonance, a topic that will be described in detail later.
Figure 53 shows an inductor and a capacitor connected in parallel. The admittance of this network is presented by equation (322) and it has only an imaginary component. If the net susceptance is positive, the susceptance of admittance is capacitive , as shown by equation (323). If the net susceptance is negative, the susceptance of admittance is inductive, as shown by equation (324).
The impedance of the parallel LC circuit is the reciprocal of its admittance, as expressed by equation (325). When w2 equals the reciprocal of the product of inductance and capacitance, the admittance is zero and the impedance is infinite, as expressed by equation (326). At this frequency, the network consisting of the inductor and the capacitor in parallel is an open circuit, as viewed from its terminals. This condition leads to resonance, a topic that will be described in detail later.
This page has described how ac circuits containing two of the three passive elements in parallel can be analyzed. The next page shows some worked examples.