SERIES RL CIRCUITS (1)

Transcription

1 SEIES IUIS () ircuit above is a series network connected to an ac voltage source Need to find the hasor form of the total imedance of this combination he total imedance of this series combination is he magnitude and angle of can be found by converting to olar form: = [ +() ] and = tan - (/) he lot of : V sin E t E t e ohms X Im e tan

2 SEIES IUIS () Examle: For a series combination circuit, = 3, =.H and e(t) = 7sin(t) V. Find the total equivalent imedance in olar form and rectangular form. Sketch the imedance in the comlex lane. We can use Ohm s law to find the total current sulied by a voltage source: i = v / Mathematical oeration must be carried out using hasors since all quantities have both magnitude and angle he current i in a series circuit is the same through every series-connected comonent he ac voltage dro across each comonent can be found by multilying each imedance by the current v = i, v = i, Examle: In a series circuit, where e(t) = 3 V, =, X =. Find the total current in the circuit. Find the voltage dros across and. Verify KV around the circuit. Draw a hasor diagram showing e, v, v and i. Sketch the voltage waveforms.

3 3 SEIES IUIS () Above is a series network connected to an ac voltage he total imedance of this combination is he olar coordinates are he hasor diagram of the total imedance is V sin E t E t e X tan tan ohms Im e -/ tan

4 SEIES IUIS () As in the series circuit, the total current sulied to the network and the voltage dros can be found by using Ohm s aw: i = e / v = i and v = i X Examle: For a series circuit where e(t) = 8sin(4t + 45) V, = 3.3k and =.µf: a) Find the total current in the circuit in hasor and sinusoidal form b) Find the voltage dros across the resistor and caacitor in hasor and sinusoidal form c) Verify KV around the circuit d) Draw a hasor diagram showing e, i, v and v e) Sketch the waveforms of e, v and v versus angle 4

5 5 SEIES IUIS () he total imedance of the circuit is In terms of magnitudes it is: = + ( X - X ) Inductive and caacitive reactance have oosite signs hus net reactance may be either inductive or caacitive, deending which is larger Polar coordinates are V sin E t E t e i(t) ohms X X X X tan tan

6 SEIES IUIS () Im he hasor diagram of the imedance when inductive reactance is greater than the caacitive reactance, i.e. when X > X / Net reactance -/ e Im he hasor diagram of the imedance when caacitive reactance is greater than the inductive reactance, i.e. when X > X / Net reactance e -/ 6

7 SEIES IUIS (3) When there is more than one resistor, caacitor and/or inductor in a series circuit, the total imedance has a resistance comonent equal to the sum of the resistance values and a reactive comonent equal to the sum of the caacitive reactances subtracted from the sum of the inductive reactances Examle: In a series circuit, e(t) = sint, = 8, = 5 and = -45. a) Find the current in olar form b) Find the voltage dros v, v, v c) Verify KV around the circuit d) Draw a hasor diagram showing e, i, v, v, v 7

8 ADMIANE he recirocal of resistance is conductance, with units Siemens he recirocal of imedance is admittance, denoted Y Y = / siemens Imedance is a measure of the extent to which a comonent imedes the flow of ac current through it Admittance is a measure of how well it admits the flow of ac current he greater the admittance, the smaller the imedance, and vice versa esistance is one form of imedance, conductance G is one form of admittance Y Phasor form of conductance is: G G siemens S 8

9 SUSEPANE eactance is another secial case of imedance he recirocal of reactance is called suscetance, B B = /X siemens here are two tyes of suscetance Inductive suscetance: B B X 9 S 9 siemens aacitive suscetance: B B X 9 S 9 siemens = /Y; = /G; X = /B ; X = /B Examle: Find the admittance of a 5 resistor; a 5mH inductor at f = 6Hz; a.µf caacitor at =.5 6 rad/s. Examle: Find the admittance Y corresonding to = Draw a hasor diagram showing Y in the comlex lane. 9

10 PAAE A IUIS () + Y - Above shows a arallel connected set of imedances to an ac source otal admittance of the circuit is the sum of the admittances of the arallel connected comonents, i.e. Y = Y + Y + + Y n = / + / + + / n Examle: Find the total admittance of a mh inductor, a k resistor and a.6µf caacitor connected in arallel to an ac voltage source of 5sin(5 3 t) V. Draw a hasor diagram showing the total admittance in the comlex lane. What is the total admittance if the frequency of the voltage source is doubled?

11 PAAE A IUIS () he total imedance of a network is the recirocal of its total admittance, = /Y For a arallel network: Y Y Y Y n n Examle: hree comonents (an inductor with imedance, a resistor with imedance, and another inductor with imedance 5) are connected in arallel. Find the total equivalent imedance of this network. Examle: A caacitor with imedance and an inductor with imedance 5 are connected in arallel. Find this network s total equivalent imedance.

12 PAAE A IUIS (3) he same voltage aears across every arallelconnected imedance So current can be found using Ohm s law and K i = i + i + + i n + i = i = i n = e/ e/ e/ n e - n i i i i n e e i e n i ey ey n i ey n

13 PAAE A IUIS (4) Examle: For the circuit below, find the current in each imedance. Show that i = ey. Draw a hasor diagram showing e, i, i, i and i 3. i e = 6sin(t) V + - i i i 3 = 5 = 4 3 = -8 3

14 A UEN SOUES () ike a dc current source, an ideal ac current source sulies the same constant current to whatever network is connected across its terminals For ac, the current is constant in the sense that its eak value does not change he same symbol is used for an ac current source as for a dc current source he direction of the arrow is ust a hase reference, since current reverses direction every half cycle eversing the arrow is the same as multilying the current by -, which is the same as adding or subtracting 8 from its hase angle I sin(t + ) A = I = -I = I +8 4

15 A UEN SOUES () Examle: For the circuit below, find the voltage v across the ac current source. Find the currents i and i. Show that i = i + i. Draw a hasor diagram showing i, v, i and i. Sketch the waveforms of i, i and i versus angle. i(t) =.4sin(5 6 t) A v i i F 5

16 POWE IN IUIS ONAINING EAANE () ecall that the average ower dissiated by a resistance carrying sinusoidal ac current can be found by P avg I V VI Veff Ieff Veff I eff Above can be used to find P avg dissiated by a resistor in a circuit containing reactance, rovided the values used in the comutations are indeed those of the voltage across and/or current through the resistor itself In many ractical circuits, it is sometimes necessary to comute P avg when the resistance or resistor voltage/current are not known 6

17 POWE IN IUIS ONAINING EAANE () i = (E / ) + e = E Above is an ac circuit with a general imedance Assuming that the voltage sulied has angle E φ E i φ Angle between voltage and current is - ( - ) = he angle between the voltage alied to a network and the total current sulied to it equals the angle of the imedance of the network esistive comonent of imedance is = cos ohms Since only resistance in network dissiates ower I I cos Pavg I E, I I E P avg I E cos 7

18 POWE FAO Equations on revious slide rovide a means for comuting the average ower in terms of voltage e across whole network and total suly current i os is called the ower factor For a urely resistive network, voltage and current are in hase, so ower factor = cos = P avg = E I / For a urely reactive network, voltage and current are searated by ±9, and cos ±9 = P avg =, i.e no ower is dissiated by urely reactive comonents (caacitors and inductors) Since V = V eff and I = I eff we can derive P avg V V I eff I eff eff cos eff cos 8

19 EXAMPE OF POWE IN A IUIS i 8 + e = 6 - In the circuit above: Find the ower factor Use the ower factor to find the average ower dissiated in the network Verify that the ower comuted above is the same as the ower comuted by using the voltage across and current through the resistor 9