Handout 11: AC circuit. AC generator

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1 Handout : AC circuit AC generator Figure compares the voltage across the directcurrent (DC) generator and that across the alternatingcurrent (AC) generator For DC generator, the voltage is constant For AC type, the voltage takes a periodic form, varying in size and sign The standard form of the AC voltage is a sinusoidal wave Assume that the amplitude is V 0 and angular frequency is ω The voltage V is given by V = V 0 cos ωt or V = V 0 sin ωt Figure : Comparison between DC voltage and AC voltage The function will be cosine or sine, depending on the initial condition However, both forms are contained in a single expression V = V 0 e iωt The cosine form is the real part of the above equation while the sine form is the imaginary part We can write down similar expression for AC current, oscillating with amplitude I 0 and angular frequency ω: Complex Impedance I = I 0 e iωt An AC generator is connected, one at a time, to a resistor, an inductor and a capacitor The generator produces current I = I 0 e iωt Resistor Consider the circuit in Fig The voltage across the resistor V = IR = I 0 R e iωt The voltage varies the same angular frequency as current and there is no phase difference (Fig 4(b)) We define complex impedance for the resistor Figure : Voltage and current across resistor Z R = V I = R The impedance is the resistance itself which is real

2 Inductor Consider the circuit in Fig 3(a) The voltage across the inductor is V = L di dt = iωli 0e iωt (iωt +π ) = ωli 0 e The voltage varies with the same frequency as current but with the phase π ahead of the current We define a complex impedance for the inductor Z L = V I = iωl The unit of the impedance is ohm (a) (b) Figure 3: Voltage and current across (a) inductor and (b) capacitor Capacitor Consider the circuit in Fig 3(b) The charge stored in the capacitor varies as Q = Idt = I 0 iω e iωt The voltage across the capacitor is V = Q C = I 0 iωc eiωt = I 0 ωc e(iωt π ) The voltage varies with the same frequency as current but with the phase π lag behind current We define a complex impedance for the capacitor Series circuit Z C = V I = iωc Consider a series circuit in Fig 4 which consists of a resistor, an inductor and a capacitor The total impedance of the circuit Z = Z R + Z L + Z C is given by Z = R + iωl + iωc = R + i ωl ωc Note the the total impedance varies with angular frequency ω Assume that the AC generator produces a varying voltage V = V 0 e iωt The current in the circuit is given by Figure 4: Series circuit I = V Z = V 0 e iωt R + i ωl ωc The above equation can be simplified by writing Z in the polar form Z = Z e iθ,

3 3 where Z = R + ωl ωc, θ = tan ωl ωc R Hence, the expression for the current in the circuit takes the form I = V 0e iωt Z e iθ = V 0e i(ωt θ) Z Figure 5: Variation of Z and phase angle θ with angular frequency ω It can be seen the amplitude of the current I = V 0 Z and the phase of current differs from that of voltage by θ Note the magnitude of the impedance is minimum when the imaginary part of Z is zero, ie, ω = LC This is known as the resonance angular frequency ω 0 The variation of total impedance with ω is shown in Fig 5 At resonance, Z = R and Z is minimum here At low frequency, θ = π and at high frequency, θ = π Because Z is minimum at the resonance, the amplitude of the current I becomes maximum here (Fig 6) Figure 6: Variation of current amplitude with ω Parallel circuit Consider a parallel circuit in Fig 7 The total impedance Z satisfies Z = + + = Z R Z L Z C R + iωl + iωc = R + i ωc ωl Assume that the AC generator produces voltage V = V 0 e iωt The total current is the circuit is given by I = V Z = V 0 R + i ωc ωl eiωt Figure 7: Parallel circuit The above can be simplified by writing Z as where Z = Z eiθ, Z = R + ωc ωl, θ = tan ωc ωl R

4 4 Hence, I = V 0 eiθ Z e iωt = V i(ωt +θ) 0e Z The amplitude of the current is I = V 0 Z and the phase of current differs from that of voltage by θ At a particular angular frequency ω = LC, the reciprocal Z is minimum Therefore, the current amplitude is also minimum at this frequency Example A series RLC circuit with L = 60 mh, C = 00 µf, and R = 400 Ω is connected to a sinusoidal voltage V volts At time t seconds, the voltage V t = 40e iωt, with ω = 00 rad s - a) What is the impedance of the circuit? b) Find the amplitude of the current in the circuit c) What is the phase difference between the voltage and the current of the circuit? Which one leads? Example An inductor with inductance L is connected in series to a resistor with resistance R The input voltage takes the form V in = V 0 e iωt The output voltage is taken across the resistor a) Show that V out = V 0 + ω L b) Sketch V out against ω Explain why this circuit is called a low-pass filter R

5 *Example 3 Consider the bridge circuit shown below When the circuit is balanced, show that 5 ω = R R 4 C C 4 Power Let V be the voltage across a device and I be the current passing through it The electrical power P is given by P = IV ( ) The sign of P varies We are interested in the average power P = IV The complex product of I and V cannot be used to find average because it is nonlinear To find the average, we must first take the real part, multiply and find the average The voltage is given by V = V 0 e iωt The real part of the voltage Re V = V 0 cos ωt The impedance of the device takes the form Z = Z e iθ Hence the current through the device is I = V Z = V 0 Z e i(ωt θ) with amplitude I 0 = V 0 Z The real part of the current Re I = V 0 Z cos ωt θ Hence, the power becomes P = V 0 Z cos ωt cos ωt θ To find P, we need to find the average cos ωt cos ωt θ = cos θ Then, the time average of power is given by P = V 0 Z cos θ For a pure capacitance or pure inductance, Z is pure imaginary, so θ = ±π/, so the average power is zero That means that no energy is dissipated in those

6 circuit elements They store energy but they do not dissipate it For a pure resistance Z = R is real, so θ = 0, so the average power is P = V 0 RThis may not immediately look like the usual relation for DC circuits, P = V R, but it is in fact equivalent, since the average value of V is just V 0 By introducing the root mean square voltage 6 V rms = V = V 0, the average power in equation ( ) can be written as P = V rms Z cos θ Similarly, we define root mean square current I rms = I = I 0, where I 0 is the amplitude of the current The rms voltage and current are the quantities usually referred to for AC circuits, rather than the amplitudes themselves which are a factor of larger Example 4 A light bulb uses average power of 70 W when connected to a 60-Hz power source with a peak voltage 70 V Calculate the resistance of the light bulb Example 5 Consider a series RLC circuit with R = 5 Ω, L = 60 mh and C = 5 μf The circuit is connected to a 600-Hz AC source with the root-mean-square voltage V rms = 0 V Calculate the average power delivered to the circuit

7 7 Supplementary note: complex number Complex number A complex number z takes the form z = a + ib where a and b are real numbers and i = The value a is called the real part and b is called the imaginary part of the complex number z Let z = a + ib and z = a + ib are complex numbers Zero complex number z = 0 means that a = 0 and b = 0 Equal complex numbes If z = z, this implies that a = a and b = b Complex conjugate The complex conjugate of z is denoted by z = a ib Addition/Subtraction z + z = a + a + i b + b z z = a a + i b b For example, 3 i + i = 3i Complex modulus A complex modulus z is defined by the equation z = zz = a + ib a ib = a + b Therefore, the complex modulus z = a + b Some useful properties of complex modulus are z z = z z z z = z z Division Multiply and divide by complex conjugate of the denominator For example, 3 i + i = 3 i + i i i = 7i 5 = i Polar form Figure shows a complex number z = a + ib plotted on a complex plane The horizontal axis is real part The vertical axis is the imaginary part

8 The vector makes angle θ to the real axis One can write a = r cos θ and b = r sin θ Hence, z = a + ib = r cos θ + ir sin θ = r cos θ + i sin θ The above is known as polar form of complex number From Fig, r = a + b = z By using Euler s formula 8 e iθ = cos θ + i sin θ, one can express z in the form Figure : Complex number z plotted on a complex plane called Argand diagram z = z e iθ, where z = a + b and θ = tan b a Note that e iθ = Example Evaluate a) 3 + i b) i c) + i ( + i) Example z = z e iθ Express each of the followings in the form a) z = + i b) z = + i 3 iπ / c) z = + i e

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