ac circuits with coupled inductors - clarkson universityjsvoboda/eta/acworkout/coils.pdfac circuits...

AC Circuits with Coupled Inductors

Introduction

The circuits in this problem set contain coupled inductors. Each problem involves the steady state response of such a circuit to a single sinusoidal input. That input is either the voltage of an independent voltage source or the current of an independent current source. Circuit analysis in the frequency-domain provides the solutions to these problems. Coupled inductors are described in Section 11.9 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda. In particular, Table 11.14-1 summarizes the equations that represent coupled inductors in the frequency domain. Circuit analysis in the frequency domain is described in Sections 10.6 thru 10.11. Table 10.7-1 summarizes the correspondence between the time domain and the frequency domain.

Worked Examples

Example 1: Consider the circuit shown in Figure 1. The input to the circuit is the current of the current source, is(t). The output is the voltage across the right-hand coil, vo(t). Determine the steady-state output voltage, vo(t).

Figure 1 The circuit considered in Example 1.

Solution: The input current is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input current. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances. Figure 2 shows the frequency domain representation of the circuit from Figure 1.

1

Figure 2 The circuit from Figure 1, represented in the frequency domain, using impedances and phasors.

The circuit in Figure 2 consists of a single mesh. The mesh current is equal to the current source Is(ω). The current Is(ω) enters the dotted end of both coils, consequently the voltage across the right hand coil is given by

( ) ( ) ( ) ( ) ( )( )(

10 5 15 15 0.707 47

15 90 0.707 47(15)(0.707) (90 47)10.605 137 V

j j j jω ω ω ω= + = = ∠ °

)= ∠ ° ∠ °

= ∠ += ∠ °

o s s sV I I I

°

In the time domain, the output voltage is given by

( ) ( )10.6 cos 5 137 Vov t t= + °



Solution: The circuit shown in Figure 3 is very similar to the circuit shown in Figure 1. There is only one difference: the dot of the right-hand coil is located at the bottom of the coil in Figure 3 and at the top of the coil in Figure 1. As in Example 1, our first step is to represent the circuit in

2

the frequency domain, using phasors and impedances. Figure 4 shows the frequency domain representation of the circuit from Figure 3.


The circuit in Figure 4 consists of a single mesh. The mesh current is equal to the current source Is(ω). The current Is(ω) enters the dotted end of the left-hand coil and enters the undotted end of the right-hand coil. Consequently the voltage across the right-hand coil is given by

( ) ( ) ( ) ( ) ( )( )(

10 5 5 5 0.707 47

5 90 0.707 47(5)(0.707) (90 47)3.535 137 V

j j j jω ω ω ω= − = = ∠ °

)= ∠ ° ∠ °

= ∠ += ∠ °

o s s sV I I I

°


( ) ( )3.535 cos 5 137 Vov t t= + °



3

Solution: The input current is sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input current. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances. Figure 6 shows the frequency domain representation of the circuit from Figure 5.


The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled in Figure 6. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that one coil current, I1(ω), enters the undotted end of its coil while the other coil current, I2(ω), enters the dotted of its coil. The device equations for coupled coils are: ( ) ( ) ( )216 8j jω ω= −1 1V I I ω (1) and ( ) ( ) ( )2 8 12j j 2ω ω= − +1V I I ω (2) The coils are connected in parallel, consequently ( ) ( )1 2ω ω= VV . Equating the expressions for V1(ω) and V2(ω) in Equations 1 and 2 gives

( ) ( ) ( ) ( )2 216 8 8 12j j j jω ω ω− = − +1 1I I I I ω

( ) ( )224 20j jω ω=1I I

( ) ( )256

ω ω=1I I

Apply Kirchhoff’s Current Law (KCL) to the top node of the coils to get

4

( ) ( ) ( ) ( ) ( ) ( )s 2 2 25 10.121 1406 6 2

1ω ω ω ω ω∠ ° = = + = + =1I I I I I I ω

( ) ( )26 0.121 140 0.066 140 A

11ω = ∠ ° = ∠I °

Now the output voltage can be calculated using Equation 2:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( )( )( )

2 2

2 2

2

8 12

8 5 12 658 126 6

5.333

5.333 90 0.066 140 0.352 230 V

j j

j j j

j

ω ω ω ω

2ω ω ω

ω

= = − +

− + = − + =

=

= ∠ ° ∠ ° = ∠ °

o 1V V I I

I I

I

I


( ) ( )0.352 cos 4 230 Vov t t= + °



Solution: The circuit shown in Figure 7 is very similar to the circuit shown in Figure 5. There is only one difference: the dot of the left-hand coil is located at the bottom of the coil in Figure 5 and at the top of the coil in Figure 7. As in Example 3, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 8 shows the frequency domain representation of the circuit from Figure 7.

5


The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled in Figure 8. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that the coil currents, I1(ω) and I2(ω), both enter the dotted of their respective coils. The device equations for coupled coils are: ( ) ( ) ( )216 8j jω ω= +1 1V I I ω (3) and ( ) ( ) ( )2 8 12j j 2ω ω= +1V I I ω (4) The coils are connected in parallel, consequently ( ) ( )1 2ω ω= VV . Equating the expressions for V1(ω) and V2(ω) in Equations 3 and 4 gives

( ) ( ) ( ) ( )2 216 8 8 12j j j jω ω ω+ = +1 1I I I I ω

( ) ( )28 4j jω ω=1I I

( ) ( )212

ω ω=1I I


( ) ( ) ( ) ( ) ( ) ( )s 2 2 21 30.121 1402 2 2ω ω ω ω ω∠ ° = = + = + =1I I I I I I ω

( ) ( )22 0.121 140 0.08067 140 A3

ω = ∠ ° = ∠I °


6

( ) ( ) ( ) ( )

( ) ( )

( )( )( )

2 2

2 2

2

8 12

18 122

16

16 90 0.04033 140 1.291 230 V

j j

j j

j

ω ω ω ω

ω ω

ω

= = +

= +

=

= ∠ ° ∠ ° = ∠ °

o 1V V I I

I I

I


( ) ( )1.291 cos 4 230 Vov t t= + °

Example 5: Consider the circuit shown in Figure 9. The input to the circuit is the voltage of the voltage source, vs(t). The output is the voltage across the right-hand coil, vo(t). Determine the steady-state output voltage, vo(t).


Solution: The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances. Figure 10 shows the frequency domain representation of the circuit from Figure 9.


7

Notice that the current, I2(ω), in the right-hand coil is zero and that both coil currents, I1(ω) and I2(ω), enter the undotted ends of their respective coils. The voltage, V1(ω),across the left-hand coil is given by ( ) ( ) ( ) ( )25 15 0 25j j jω ω= + =1 1V I I ω1 (5) Apply KVL to the mesh consisting of the voltage source, resistor and the left-hand coil to get

( ) ( )8 7.32ω ω 95 0+ − ∠ ° =1 1I V Substituting the expression for V1(ω) from Equation 5 gives

( ) ( )8 25 7.32 95jω ω 0+ − ∠ ° =1 1I I Now solve for I1(ω) to get

( ) 7.32 95 7.32 95 0.279 23 V8 25 26.25 72j

ω ∠ ° ∠ °= = = ∠

+ ∠ °1I °

The voltage, Vo(ω), across the right-hand coil is given by

( ) ( ) ( ) ( ) ( )( )

15 10 0 15 15 0.279 23

(15 90 ) 0.279 234.185 113 V

j j j jω ω ω= + = = ∠ °

= ∠ ° ∠ °

= ∠ °

o 1 1V I I


( ) ( )4.185 cos 4 113 Vov t t= + °

8



Solution: The circuit shown in Figure 11 is very similar to the circuit shown in Figure 9. There is only one difference: the dot of the right-hand coil is located at the bottom of the coil in Figure 9 and at the top of the coil in Figure 11. As in Example 5, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 12 shows the frequency domain representation of the circuit from Figure 11.


Notice that the current, I2(ω), in the right-hand coil is zero and that one of coil currents, I1(ω), enters the undotted end of the coil while the other coil current, I2(ω), enters the dotted end of the coil. The voltage, V1(ω),across the left-hand coil is given by ( ) ( ) ( ) ( )25 15 0 25j j jω ω= − =1 1V I I ω1 (6) Apply KVL to the mesh consisting of the voltage source, resistor and the left-hand coil to get

( ) ( )8 7.32ω ω 95 0+ − ∠ ° =1 1I V

9

Substituting the expression for V1(ω) from Equation 6 gives

( ) ( )8 25 7.32 95jω ω 0+ − ∠ ° =1 1I I Now solve for I1(ω) to get

( ) 7.32 95 7.32 95 0.279 23 V8 25 26.25 72j

ω ∠ ° ∠ °= = = ∠

+ ∠ °1I °

The voltage, Vo(ω), across the right-hand coil is given by

( ) ( ) ( ) ( ) ( )( )

15 10 0 15 15 0.279 23

(15 90 ) 0.279 23 4.185 67 V

j j j jω ω ω= − + = − = − ∠ °

= ∠− ° ∠ ° = ∠− °o 1 1V I I


( ) ( )4.185 cos 4 67 Vov t t= − °



Solution: The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances. Figure 14 shows the frequency domain representation of the circuit from Figure 13.

10


The circuit in Figure 14 consists of a single mesh. Notice that the mesh current, I(ω), enters the undotted ends of both coils. Apply KVL to the mesh to get

( ) ( ) ( )( ) ( ) ( )( )5 12 6 6 15 5.94 140 0j j j jω ω ω ω ω+ + + + − ∠I I I I I ° =

( ) ( ) ( )5 12 6 6 15 5.94 140 0j j j jω ω+ + + + − ∠ ° =I I

( ) ( )5.94 140 5.94 140 5.94 140 0.151 57 A

5 12 6 6 15 5 39 39.3 83j jω ∠ ° ∠ ° ∠ °

= = = =+ + + + + ∠

I ∠ °

Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to the passive convention. The voltage across the right-hand coil is given by is given by

( ) ( ) ( ) ( ) ( )( )

15 6 21 21 0.151 57

(21 90 ) 0.151 573.17 147 V

j j j jω ω ω ω= + = = ∠ °

= ∠ ° ∠ °

= ∠ °

oV I I I


( ) ( )3.17 cos 3 147 Vov t t= + °

11



Solution: The circuit shown in Figure 15 is very similar to the circuit shown in Figure 13. There is only one difference: the dot of the left-hand coil is located at the right of the coil in Figure 13 and at the left of the coil in Figure 15. As in Example 7, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 16 shows the frequency domain representation of the circuit from Figure 15.


The circuit in Figure 14 consists of a single mesh. Notice that the mesh current, I(ω), enters the dotted end of the left-hand coil and the undotted end of the right-hand coil. Apply KVL to the mesh to get

( ) ( ) ( )( ) ( ) ( )( )5 12 6 6 15 5.94 140 0j j j jω ω ω ω ω+ − + − + − ∠I I I I I ° =

( ) ( ) ( )5 12 6 6 15 5.94 140 0j j j jω ω+ − − + − ∠ ° =I I

( ) ( )5.94 140 5.94 140 5.94 140 0.376 68.4 A

5 12 6 6 15 5 15 15.8 71.6j jω ∠ ° ∠ ° ∠ °

= = = =+ − − + + ∠

I ∠ °

12

Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to the passive convention. The voltage across the right-hand coil is given by is given by

( ) ( ) ( ) ( ) ( )( )

15 6 9 9 0.376 68.4

(9 90 ) 0.376 68.43.38 158.4 V

j j j jω ω ω ω= − = = ∠ °

= ∠ ° ∠ °

= ∠ °

oV I I I


( ) ( )3.38 cos 3 158.4 Vov t t= + °



Solution: The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances. Figure 18 shows the frequency domain representation of the circuit from Figure 17. The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled in Figure 18. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that both coil currents, I1(ω) and I2(ω), enter the undotted ends of their respective coils.

13


The device equations for coupled coils are: ( ) ( ) ( )216 8j jω ω= +1 1V I I ω (7) and ( ) ( ) ( )2 8 20j j 2ω ω= +1V I I ω (8) The coils are connected in parallel, consequently ( ) ( )1 2ω ω= VV . Equating the expressions for V1(ω) and V2(ω) in Equations 7 and 8 gives

( ) ( ) ( ) ( )2 216 8 8 20j j j jω ω ω+ = +1 1I I I I ω

( ) ( )28 12j jω ω=1I I

( ) ( )232

ω ω=1I I


( ) ( ) ( ) ( ) ( ) ( )2 2 2 23 52 2

ω ω ω ω ω= + = + =1I I I I I I ω

Therefore

( ) ( ) ( ) ( )23 and 5 5

2ω ω ω=1I I I I ω= (9)

Substituting the expressions for I1(ω) and I2(ω) from Equation 9 into Equation 7 gives

14

( ) ( ) ( ) ( ) ( ) ( ) ( )16 3 8 23 216 8 12.85 5 5

j j j jω ω ω ω+ = + = =

1V I I I ωI (10)

Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get

( ) ( )4 5.7 158 0ω ω+ − ∠ ° =1I V Using Equation 10 gives

( ) ( )4 12.8 5.7 158 0jω ω+ − ∠ ° =I I Solving for I (ω) gives

( ) 5.7 158 5.7 158 0.425 85 A4 12.8 13.41 73j

ω ∠ ° ∠ °= = = ∠

+ ∠ °I °


( ) ( ) ( )

( )( )( )

12.8

12.8 0.425 85

12.8 90 0.425 85 5.44 175 V

j

j

ω ω ω= =

= ∠ °

= ∠ ° ∠ ° = ∠ °

o 1V V I


( ) ( )5.44 cos 4 175 Vov t t= + °



15

Solution: The circuit shown in Figure 19 is very similar to the circuit shown in Figure 17. There is only one difference: the dot of the right-hand coil is located at the bottom of the coil in Figure 17 and at the top of the coil in Figure 19. Figure 20 shows the frequency domain representation of the circuit from Figure 19.


The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled in Figure 20. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that one of coil currents, I1(ω), enters the undotted end of the coil while the other coil current, I2(ω), enters the dotted end of the coil. The device equations for coupled coils are: ( ) ( ) ( )216 8j jω ω= −1 1V I I ω (11) and ( ) ( ) ( )2 8 20j j 2ω ω= − +1V I I ω (12) The coils are connected in parallel, consequently ( ) ( )1 2ω ω= VV . Equating the expressions for V1(ω) and V2(ω) in Equations 11 and 12 gives

( ) ( ) ( ) ( )2 216 8 8 20j j j jω ω ω− = − +1 1I I I I ω

( ) ( )224 28j jω ω=1I I

( ) ( ) ( )2 228 724 6

ω ω ω= =1I I I


16

( ) ( ) ( ) ( ) ( ) ( )2 2 2 27 16 6

3ω ω ω ω ω= + = + =1I I I I I I ω

Therefore

( ) ( ) ( ) ( )27 and

13 136ω ω ω=1I I I I ω= (13)

Substituting the expressions for I1(ω) and I2(ω) from Equation 13 into Equation 11 gives

( ) ( ) ( ) ( ) ( ) ( ) ( )16 7 8 67 616 8 4.913 13 13

j j j jω ω ω ω− = − = =

1V I I I ωI (14)

Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get

( ) ( )4 5.7 158 0ω ω+ − ∠ ° =1I V Using Equation 14 gives

( ) ( )4 4.9 5.7 158 0jω ω+ − ∠ ° =I I Solving for I (ω) gives

( ) 5.7 158 5.7 158 0.901 107 A4 4.9 6.325 51j

ω ∠ ° ∠ °= = = ∠

+ ∠ °I °


( ) ( ) ( )

( )( )( )

4.9

4.9 0.901 107

4.9 90 0.901 107 4.41 197 V

j

j

ω ω ω= =

= ∠ °

= ∠ ° ∠ ° = ∠ °

o 1V V I


( ) ( )4.41 cos 4 197 Vov t t= + °

17

ac circuits with coupled inductors - clarkson universityjsvoboda/eta/acworkout/coils.pdfac circuits...

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