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Simple AC Circuits

Introduction

Each problem in this problem set involves the steady state response of a linear, time-invariantcircuit to a single sinusoidal input. Such a response is known to be sinusoidal and to have thesame frequency as the input. The linear, time invariant circuits in this problem set consist of resistors, capacitors and inductors. The input to each circuit is either the voltage of anindependent voltage source or the current of an independent current source.

The circuits in this problem set

1. Are linear and time-invariant2. Have a single sinusoidal input

3. Are at steady state

Consequently, these circuits can be represented in the frequency-domain, using impedances and phasors. Analysis in the frequency-domain is preferred to analysis in the time-domain becausethe frequency-domain analysis involves algebraic equations rather than differential equations.Unfortunately, solving the frequency domain equations requires algebra and arithmetic withcomplex numbers.

The steady-state response of circuits with sinusoidal inputs is the topic of Chapter 10 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda. In particular, phasors aredescribed in Section 10.6 and 10.7 and impedance is defined in Section 10.8. Table 10.7-1

summarizes the correspondence between the time domain and the frequency domain. Also,Appendix B provides a review of complex arithmetic.

Worked Examples

Example 1:Consider the circuit shown in Figure 1. The input to the circuit is the voltage of the voltagesource, vs(t ). The output is the voltage across the resistor, vo(t ). Determine inductance, L, of theinductor.

Figure 1 The circuit considered in Example 1.

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Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the samefrequency as the input voltage. Apparently the circuit has reached steady state. Consequently, thecircuit in Figure 1 can be represented in the frequency domain, using phasors and impedances.Figure 2 shows the frequency domain representation of circuit from Figure 1. The voltages V s()and Vo() in Figure 2 are the phasors corresponding to vs(t ) and vo(t ) from Figure 1. The inductor

and the resistor are represented as impedances in Figure 2. The impedance of the inductor is, as shown in Figure 2.( )1 j L j L j L = =

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

The current I () in Figure 2 is given by

( ) ( ) 2.77 307 0.396 307 A7 7

o

= = = VI

The inductor voltage, VL(), in Figure 2 is given by

( ) ( ) ( ) 4.41 178 2.77 3073.427 37

= = =

L s oV V V

The impedance of the inductor is given by

( )( )

3.427 378.65 270 8.65

0.396 307 j L j

= = = =

LV

I

Therefore L = 8.65 H.

Alternate Solution: Apply the voltage divider principle to the circuit in Figure 2 to get

( )72.27 307 4.41 1787 j L

= +

Doing some algebra gives

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( )( )7 2.27 307 2.27 307

0.628 517 4.41 178 1 180 4.41 178 j L

= = = +

Solving for L gives( )

( )7 1 0.628 51

8.66 0.013 8.65 H0.628 51 L j j

= =

Example 2:Consider the circuit shown in Figure 3. The input to the circuit is the voltage of the voltagesource, vs(t ). The output is the voltage across the capacitor, vo(t ). Determine resistance, R, of theresistor.


Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the samefrequency as the input voltage. Apparently the circuit has reached steady state. Consequently, thecircuit in Figure 3 can be represented in the frequency domain, using phasors and impedances.Figure 4 shows the frequency domain representation of circuit from Figure 3. The voltages V s()and Vo() in Figure 4 are the phasors corresponding to vs(t ) and vo(t ). The capacitor and the

resistor are represented as impedances in Figure 4. The impedance of the capacitor wascalculated as

( )1 1

0.9973 0.3342

j j j jC

= =


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( ) ( ) 1.89 230 1.89 320 A1 90

o

j

= = =

VI

The resistor voltage, VR (), in Figure 4 is given by

( ) ( ) ( ) 7.83 126 1.89 2307.597 40

= = =

R s oV V V

The impedance of the resistor is given by

( )( )

7.597 404.02 360 4

1.89 320 R

= = =

R V

I

Therefore R = 4 .


( )1.89 230 7.83 126 j j R

=

+

Doing some algebra gives

( )( )1.89 230 1.89 230

0.241 767.83 126 1 180 7.83 126

j j R

= = = +

Solving for L gives( )( )0.241 76 1

4.02 0.0022 40.241 76

j R j

= =

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Example 3:Consider the circuit shown in Figure 5. The input to the circuit is the voltage of the voltagesource, vs(t ). The output is the voltage across the inductor, vo(t ). Determine amplitude, A, of vo(t ).


Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the same

frequency as the input voltage. Apparently the circuit has reached steady state. Consequently, thecircuit in Figure 5 can be represented in the frequency domain, using phasors and impedances.Figure 6 shows the frequency domain representation of circuit from Figure 5. The voltages V s()and Vo() in Figure 6 are the phasors corresponding to vs(t ) and vo(t ). The inductor and theresistor are represented as impedances. The impedance of the inductor is

, as shown in Figure 6.( )( )4 0.54 2.16 j L j j = =


Apply the voltage divider principle to the circuit in Figure 6 to get

( )2.16311 7.28 77 4.254 49 4.254 3113 2.16

j A

j = = =

+

Therefore A = 4.254 V.

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Example 4:Consider the circuit shown in Figure 7. The input to the circuit is the voltage of the voltagesource, vs(t ). The output is the voltage across the resistor, vo(t ). Determine capacitance, C , of thecapacitor.


Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the same

frequency as the input voltage. Apparently the circuit has reached steady state. Consequently, thecircuit in Figure 7 can be represented in the frequency domain, using phasors and impedances.Figure 8 shows the frequency domain representation of circuit from Figure 7.



( ) ( ) 1.59 125 1.59 125 A1 1 0

o

= = =

VI

The capacitor voltage, VC(), in Figure 4 is given by

( ) ( ) ( ) 7.68 47 1.59 1257.512 35

= = =

C s oV V V

The impedance of the capacitor is given by

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( )

( )1 7.512 35

4.725 902 1.59 125

jC

= = =

CV

I

Solving for C gives

( )0.106 F

2 4.725 90

jC

= =


( )11.59 125 7.68 471

12

jC

=

+

Doing some algebra gives2 1.59 125

0.207 782 7.68 47

C

C j

= =

Solving for C gives

( )( )

( ) ( )( )( )

( )( )( )

0.207 78 1 90 0.207 78

2 0.207 78 1 2 0.043 0.202 1

0.207 168

2 0.957 0.202

0.207 1680.106 0 0.106

2 0.978 168

jC

j

j

j

= =

+

=

+

= = + =

Example 5:Consider the circuit shown in Figure 9. The input to the circuit is the voltage of the voltagesource, vs(t ). The output is the voltage across the capacitor, vo(t ). Determine amplitude, A, of vo(t ).


Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the samefrequency as the input voltage. Apparently the circuit has reached steady state. Consequently, the

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circuit in Figure 9 can be represented in the frequency domain, using phasors and impedances.Figure 10 shows the frequency domain representation of circuit from Figure 9. The impedance of the capacitor was calculated as

( ) ( )

1 11.07

1 0.933 j j j

C = =

Apply the voltage divider principle to the circuit in Figure 10 to get

( )1.0764 7.2 139 1.864 644 1.07

j A

j

= =

Therefore A = 1.864 V.

Figure 10 The circuit from Figure 9, represented in the frequency domain, using impedances and phasors

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simple ac

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