Simple AC Circuits. Introduction

Transcription

1 Simple AC Circuits Introduction Each problem in this problem set involves the steady state response of a linear, time-invariant circuit to a single sinusoidal input. Such a response is known to be sinusoidal and to have the same 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 an independent voltage source or the current of an independent current source. The circuits in this problem set 1. Are linear and time-invariant 2. 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 because the frequency-domain analysis involves algebraic equations rather than differential equations. Unfortunately, solving the frequency domain equations requires algebra and arithmetic with complex 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 are described in Section 10.6 and 10.7 and impedance is defined in Section Table 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 voltage source, v s (t). The output is the voltage across the resistor, v o (t). Determine inductance, L, of the inductor. Figure 1 The circuit considered in Example 1. 1

2 Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the same circuit 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 (ω) and V o (ω) in Figure 2 are the phasors corresponding to v s (t) and v o (t) from Figure 1. The inductor and the resistor are represented as impedances in Figure 2. The impedance of the inductor is jω L= j 1 L= jl, as shown in Figure 2. () Figure 2 The circuit from Figure 1, represented in the frequency domain, using impedances and phasors. The current I(ω) in Figure 2 is given by The inductor voltage, V L (ω), in Figure 2 is given by Vo I = = = A 7 7 ( ω) ( ω) ( ω) V = V V = L s o The impedance of the inductor is given by Therefore L = 8.65 H. V L = jl= = = = j8.65 I Alternate Solution: Apply the voltage divider principle to the circuit in Figure 2 to get Doing some algebra gives = jl 2

3 = = = jl Solving for L gives ( ) j ( ) L= = 8.66 j H Example 2: Consider the circuit shown in Figure 3. The input to the circuit is the voltage of the voltage source, v s (t). The output is the voltage across the capacitor, v o (t). Determine resistance, R, of the resistor. Figure 3 The circuit considered in Example 2. Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the same circuit 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 (ω) and V o (ω) in Figure 4 are the phasors corresponding to v s (t) and v o (t). The capacitor and the resistor are represented as impedances in Figure 4. The impedance of the capacitor was calculated as 1 1 j = j = j j Ω ω C Figure 4 The circuit from Figure 3, represented in the frequency domain, using impedances and phasors. 3

4 The current I(ω) in Figure 3 is given by The resistor voltage, V R (ω), in Figure 4 is given by Vo I = = = A j 1 90 ( ω) ( ω) ( ω) V = V V = R s o = The impedance of the resistor is given by V R R = = = I Therefore R = 4 Ω. Alternate Solution: Apply the voltage divider principle to the circuit in Figure 4 to get Doing some algebra gives j = j+ R j = = = j+ R Solving for L gives (( ) 1) j R= = 4.02 j Ω 4

5 Example 3: Consider the circuit shown in Figure 5. The input to the circuit is the voltage of the voltage source, v s (t). The output is the voltage across the inductor, v o (t). Determine amplitude, A, of v o (t). Figure 5 The circuit considered in Example 3. Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the same circuit 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 (ω) and V o (ω) in Figure 6 are the phasors corresponding to v s (t) and v o (t). The inductor and the resistor are represented as impedances. The impedance of the inductor is jω L= j = j2.16ω, as shown in Figure 6. Figure 6 The circuit from Figure 5, represented in the frequency domain, using impedances and phasors. Apply the voltage divider principle to the circuit in Figure 6 to get j 2.16 A 311 = ( ) = = j 2.16 Therefore A = V. 5

6 Example 4: Consider the circuit shown in Figure 7. The input to the circuit is the voltage of the voltage source, v s (t). The output is the voltage across the resistor, v o (t). Determine capacitance, C, of the capacitor. Figure 7 The circuit considered in Example 4. Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the same circuit 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. Figure 8 The circuit from Figure 7, represented in the frequency domain, using impedances and phasors. The current I(ω) in Figure 3 is given by Vo I = = = A The capacitor voltage, V C (ω), in Figure 4 is given by ( ω) = ( ω) ( ω) = V V V C s o = The impedance of the capacitor is given by 6

7 Solving for C gives V C j = = = C I j C = = F ( ) Alternate Solution: Apply the voltage divider principle to the circuit in Figure 8 to get Doing some algebra gives Solving for C gives = j 2 C 2 C = = C j j ( j ) ( ) 2 ( j0.202) ( ) 2 ( ) C = = = = = j0 = Ω Example 5: Consider the circuit shown in Figure 9. The input to the circuit is the voltage of the voltage source, v s (t). The output is the voltage across the capacitor, v o (t). Determine amplitude, A, of v o (t). Figure 9 The circuit considered in Example 5. Solution: The input voltage is a sinusoid. The output voltage is also a sinusoid and has the same 7

8 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 1 j = j = j1.07 Ω ω C ()( ) Apply the voltage divider principle to the circuit in Figure 10 to get Therefore A = V. j 1.07 A 64 = ( ) = j 1.07 Figure 10 The circuit from Figure 9, represented in the frequency domain, using impedances and phasors 8