www.sakshieducation.com ELECTROMAGNETICS 6. AC CIRCUITS POINTS TO REMEMBER 1. The electric current which does not change its magnitude or direction with time is known as direct current. 2. The electric current whose magnitude changes with time and direction reverses periodically is known as alternating current. ∴ e = e 0 sin ωt Where e 0 = BAN ω is known as peak value of alternating emf Also, e = e 0 sin ωt and i = i0 sin ωt Where i0 = peak value of alternating current. 3. Average or mean value: a) It is the steady current (dc) which when passes through a circuit for half time period of AC sends the same charge as done by the AC in the same time. b) The average value of AC for the complete cycle is zero. c) The average value of AC for the half cycle of AC is given by. 2I I avg = 0 = 0.636I 0 π Similarly, Eavg is also 2E 0 π for half cycle of AC 4. RMS value or Virtual value or Effective value : The RMS value of AC is the steady current (DC) which when flowing through a given resistance for a given time, produces the same amount of heat as produced by the AC when flowing through the same resistance for the same time. E0 = 0.707E0 2 2 RMS value of AC for half cycle is also same as above 5. AC through a pure resistor ; E = E 0 sin ωt and I = I 0 sin ωt Where E0 and I0 are the peak values of voltage and current of AC respectively. IV = I0 = 0.707I 0 and EV = Since the emf and current raise or fall simultaneously they are in phase with each other. 6. AC through a pure inductor: E π E = E 0 sin ωt and I = 0 sin ω t − Lω 2 The term L ω has the units of resistance and it is called as resistance of the inductor or inductive reactance (XL) E π π I = 0 sin ωt − ⇒ I = I 0 sin ωt − XL 2 2 Where I0 is the peak value of current. Hence the current is lagging behind emf or emf is leading current by a phase difference of 7. AC Through a pure capacitor: E E = E 0 sin ωt and I = 0 cos ωt 1 cω π 2 . www.sakshieducation.com 9.com .com 1 is called capacitive reactance where it has the dimensions of Cω resistance.www. It is also called resistance of capacitor. then Tan φ = 0 or φ = 0 and hence emf and current are in phase.sakshieducation. VC = IX C = 1 b) E = I R 2 + Lω − Cω 2 I And VR = IR . e) If Lω = 1 . E π π I = 0 sinωt + ⇒ I = I0 sinω t + Where I0 is peak value of AC XC 2 2 The term 2 8. AC through C – R circuit : I VC = I × X C = And VR = IR Where XC = capacitive reactance Cω Z = R 2 + ( Lω ) and Tan φ = 2 E = VR + VC2 ⇒ E = ( IR ) 1 2 I + Cω 2 E = I R 2 + (C ω ) 2 and Tan φ = VC 1 = VR CωR This is the phase angle by which the emf lags behind the current in C – R circuit 1 Here z = R 2 + Cω 2 is called the impedance of C – R circuit.sakshieducation. c) Z = R 2 + Lω − Cω 2 d) 1 Lω − V L − VC Cω Tan φ = = VR R This is the phase angle by which the emf leads the current. AC through LR circuit : Let VL and VR be the instantaneous voltages across the inductor and resistor respectively. z is called impedance of LR circuit. www. Cω 1 is called the impedance of L – C – R circuit. AC through L – C – R circuit : a) VL = IX L = ILω . 10. ∴E = I R 2 + (L ω ) 2 2 Hence current leads emf or emf is lagging behind current by π Lω R This is the phase angle by which the emf leads the current in L – R circuit. This condition is called resonance condition. VL = IX L = IL ω And VR = IR Where XL is the inductive reactance when AC flow through a pure resistor. The Cω circuit in this condition behaves like a pure resistor circuit. Derive expression for impedance and current of an a. 12. Lω = ω = 2π n = 1 LC 1 Cω Or ω 2 = 1 1 2π LC LC ⇒ n = 11. the voltage and current are in phase unit each other which are represented along OX when AC flows through a . Since L and R are in series.sakshieducation. LONG ANSWER QUESTIONS 1. circuit containing (i) inductance and resistance in series and (ii) capacitance and resistance in series. 3.sakshieducation. If OA and OB represents the magnitudes of VR and VL then OC represents the effective value of emf in the circuit therefore. d) AC can be easily converted into DC by using rectifiers.www. AC through LR circuit : 1.com . = ∴E = I Y VL B VL ( IR ) 2 + ( IL ω ) 2 2 R 2 + (L ω ) E C φ O VR A 2 VR I X Z = R 2 + ( Lω ) www. Consider a source of alternating emf connected to a series combination of an inductor L and a resistor R. 2 + VL2 OC = OA2 + OB2 = VR pure inductor.c. Advantages of AC over DC : a) The generation of AC is more economic than DC b) AC voltages can be easily stepped up or stepped down using transformers. And VR = IR VL = IX L = IL ω Where XL is the inductive reactance when AC flow through a pure resistor. Let VL and VR be the instantaneous voltages across the inductor and resistor respectively.com f) At resonance. c) AC can be transmitted to longer distances with less loss of energy. Let E be the instantaneous emf and I be the instantaneous current in the circuit. Disadvantages : a) AC is more fatal and dangerous than DC b) AC always flows on the outer layer of the conductor (skin effect) and hence AC requires stranded wires. Ans. L VL R VR ~ 2. the current I 2 is same in both and VL is represented along OY both and VL is represented along OY. voltage leads the current by phase π 4. c) AC cannot be used in electrolysis like electroplating etc. sakshieducation.f.com Where z is called impedance of LR circuit.m. Tan φ = V C = VR This is the phase angle by which the emf lags behind the current in C – R circuit 2. Ans. www. ‘C’ and ‘L’ in series. from the diagram. Deduce an expression for the resonating frequency of an LCR series resonating circuit.m. 1 CωR Also. Consider a source of alternate emf connected to a series combination of a capacitor C and a resistance R.com . As shown in figure let ‘i’ be the current and ‘q’ be charge at any instant time ‘t’. An e. I VC = I × X C = And VR = IR Cω 2. 3. Also from the diagram OB VL IX L X L Lω = = = Tan φ = ⇒ Tan φ = OA VR IR R R This is the phase angle by which the emf leads the current in L – R circuit. q di and across inductor is L .www. Y VL O φ π 2 VR A VR I X VC B Y’ C E 2 2 OC = OA + OB 2 E = VR + VC2 ⇒ E = ( IR ) + 2 2 I + Cω 2 E = I R 2 1 (C ω ) 2 1 Here z = R 2 + Cω is called the impedance of C – R circuit. When AC flows through a pure capacitor.f e = E0 sin wt be applied to a circuit consisting of ‘R’. the emf is logging behind current by a phase which is represented along OY’.sakshieducation. Obtain an expression for impedance and current in series LCR circuit. Where XC=capacitive reactance When AC flows through a pure resistor. the voltage and current are in phase with each other which are represented along OX. C– R circuit : 1. both opposite The potential difference across capacitor is C dt applied e. which is at low frequencies ωC I 0 is maximum current given by E0 I0 = 2 1 R2 + ω L − ωC E 1 Impendence ( Z ) = 0 = R 2 + ω L − ωC I0 At a particular frequency. iii) Root mean square (R.www. ω2 = 1 =0 ωC 1 LC SHORT ANSWER QUESTIONS 1. This frequency is called resonant frequency ( f 0 ) .e. At resonance. I = I 0 sin ωt ii) In the above equation I 0 is known maximum current. this net e. we have i = i0 sin(ωt + φ ) If ω L > ωC 1 If ω L < .m. di q ∴ E0 sin ωt − L − = Ri dt C di q We have L + Ri + = E0 sin ωt dt C The current ‘i’ is given by i = i0 sin(ωt − φ ) 1 .S) value: www.com Hence net e. impedance of series LCR circuit is minimum and is equal to R. ω L − = 0 under this condition impedance Z = R ωC 1 ∴ Z = R2 + ω L − ωC 1 ∴ Z = R ∴ ω L = ωC At this frequency. and rms values of a current. at this frequency current is maximum. i) The strength of the current at any instant of time‘t’ is known as instantaneous current. the total reactance of the circuit is zero.m.com . ω L − 2 2 1 or ω = LC 1 2π f 0 = LC 1 f0 = 2π LC The frequency response curve of LCR is shown in figure.sakshieducation. It is given by. While ω is angular velocity. maximum.f is equal to iR.M. 1 i.f is Ldi q E0 sin ωt − − dt C According to Ohm’s law. Sol: Explain instantaneous.sakshieducation. R E E ∴ I = I 0 sin ωt Q I = = 0 sin ωt R R Where E0 and I0 are the peak values of voltage and current of AC respectively. Consider a pure resistance of resistance R(zero inductance) is connected to a source of an alternate emf .sakshieducation.www.f is applied. produces the same amount of heat as produced by the AC when flowing through the same resistance for the same time.m. The instantaneous emf is given by E = E 0 sin ωt dI be the rate of change Let I be the current through the circuit and dt of current in the circuit at any instant. E ∫ dI = L0 ∫ sin ωt dt E I = 0 ( − cos ωt ) Lω E π I = 0 sin ω t − 2 Lω The term L ω has the units of resistance and it is called as resistance of the inductor or inductive reactance (XL) E π π I = 0 sin ωt − ⇒ I = I 0 sin ωt − XL 2 2 3.com . Discus the flow of a. E = E 0 sin ωt The current through the resistance at any instant is given by. LdI E= dt E E or dI = dt = 0 sin ωt dt L L Integrating the above equation.com The RMS value of AC is the steady current (DC) which when flowing through a given resistance for a given time.D.c. and current in a resistor. Consider a pure inductor which has no resistance connected to a source of emf E. Sol: ~ E E T/2 3T/4 t T/4 θ = ωt 0 X Since the emf and current raise or fall simultaneously they are in phase with each other There is no phase difference between the E. e. Obtain an expression for the current in a inductance when an a. The net emf in the circuit is given by LdI LdI E− =0 ⇒ E= dt dt Since there is no resistance in the circuit there is no P. in the circuit. Y E I I E (or) I 2.The instantaneous value of alternating emf is given by. www. through a pure resistor. Ans.c.f.sakshieducation.m. f. impedance Z = R + ωC 1 is capacitive reactance ( X c ) where ωC 2 2 www.m. there is no phase difference between a.f c) In pure capacitor a. circuit. e. and current. Consider a capacitor of capacity C connected to a source of alternating emf E.com .www. Hence the current is lagging behind emf or emf is leading current by a phase difference of 4.m. value (or effective or virtual value) of A.m.c. q q E − = 0 (or) E = C C (or) q = EC = CE 0 sin ωt E dq ∴I = = CE 0ω cos ωt (or) I = 0 cos ωt 1 dt cω 1 is called capacitive reactance where it has The term Cω dimensions of resistance. current lags behind the e. E π π I = 0 sinωt + ⇒ I = I0 sinω t + 2 XC 2 Where I0 is peak value of AC ~ E the Hence current leads emf or emf is lagging behind current by π 2 VERY SHORT ANSWER QUESTIONS 1. Ans. e Here e0 = maximum or peak value of A.C. The instantaneous emf is given by. radian. Voltage (erms ) : It is the square root of the average of squares of all the of all the instantaneous values of voltage over one complete cycle. impedance Z = R 2 + ( Lω ) 2 Where Lω is inductive reactance ( X L ) 1 ii) In C-R series circuit. in one complete cycle.m.C.s value of the voltage in an A. current leads emf by 3. erms = 0 = 0.C.f.s. pure inductor and pure capacitor: a) In pure resistor a. b) In pure inductor a. Ans.com Where I0 is the peak value of current. What is the r. E = E 0 sin ωt C The net emf in the circuit is given by.m.f. Ans.m.c. What is the impedance in the following series circuits? i) L and R ii) C and R i) In L-R series circuit.c. circuit. is applied. Obtain an expression for the current in a capacitor when an a. e. circuit? r.c.707 e0 2 What is the phase difference between a.c. 2. circuit.c. π π 2 2 radian. π 2 . Ans.sakshieducation.sakshieducation. e. and current in the following? Pure resistor. It is also called resistance of capacitor. Sol: 5.7 V ω = 2π f = 2π ( 60 ) = 120π rad/s e = (169. Find the inductive reactance. e0 = 300 volt.. 2 2 Write the equation of an alternating emf of 120 V. e 300 = 3A The peak current = 0 = R 100 peak current 3 = = 2. e0 = 2 e = 2 ×120 = 169. Calculate the rms current through the circuit Comparing the alternating emf.sakshieducation. 120 V represents the rms emf i. 2 2 Comparing e = 311sin100π t with e = e0 sin ωt . A hot e0 = 100V ω = 100 π rad/s 2 π f = 100π Sol: wire instrument is used to measure the alternating voltage. The rms current= 2 2 A capacitance of 0. Sol: peak emf.7 sin120 π t ) volt.www. e = 100sin (100π t ) Sol: Comparing it with e = e0 sin π t Frequency f = 2.e =300 sin (100π ) t volt is applied to a pure resistance of 100 Ω ..e.4 ×10−6 F . f = 100 Hz 1 1 1 XC = = = = 3978Ω −6 Cω C ( 2π f ) 0.9 V.4 µ F is connected to an alternating emf of 100 cycles/second. e = 300 sin (100π ) t volt with e = e0 sin ω t . f = 50 Hz www. Sol: 6. E = . ∴ The equation of the alternating emf is e = e0 sin ω t 4.com 4. e = 120 V The peak emf.4 ×10 × 2π ×100 An inductor of inductance 2 mH is connected to an alternating emf of 50 Hz. Sol: e0 = 311 V e 311 e= 0 = = 219.com . Find the peak value of its emf and SOLVED PROBLEMS 1. State the expression for the reactance of (i) an inductor and (ii) a capacitor.121A . i) Inductive reactance ( X L ) = ω L 1 ii) Capacitive reactance ( X C ) = ωC The equation of an alternating emf is e =100 sin (100π ) t. 100π = 50 Hz 2π The equation of the alternating voltage supplied to our houses is e = ( 311sin100π t ) volt. An alternating emf. What is the capacitive reactance? C = 0. we get 3. What reading does it show? peak voltage e0 = Rms voltage.sakshieducation. 60 Hz. L = 2mH = 2 ×10−3 H . its frequency. Ans.4µ F = 0. a capacitance of 2 µ F and a resistance of 20 Ω are connected in series across an alternating emf of 220 V.7 sin (120 π t ) volt. f = 50 Hz i) The impedance of the circuit.’s across the inductor. Z 1277 iii) The peak current. a capacitance of 0.sakshieducation.d.6284Ω 5.com The inductive reactance.s.1µ F = 0. Find the peak value of its ω = 102 π rads −1 ω 120π n= = = 60 Hz 2π 2π 2.2Ω XC = 1 1 1 = = = 1591Ω −6 Cω C ( 2π f ) 2 ×10 × 2π × 50 2 2 2 Z = R 2 + ( X L − X C ) = 202 + ( 314. the p.1V A resistance of 10 Ω .m.1×10−6 2. Z = R 2 + ( X L − X C ) R= 20Ω. E = 169. Vrms = 220V ⇒ V0 = Vrms × 2 = 220 2V N = 50 Hz w = 2π n = 2π (50) = 100π rad s −1 www. 2 UNSOLVED PROBLEMS 1. rms emf. Sol: Write the equations of a sinusoidal voltage of r. capacitor and resistor are 70 V.com .1 µ F and an inductance of 2 mH are connected in series across a source of alternating emf of variable frequency. EC = 30V f0 = 1 = 2 E = ER + ( EL − EC ) .www.7 V The equation of an alternating e. X L = Lω = L ( 2π f ) = 2 ×10−3 × 2π × 50 = 0.1723 A.25kHz 2π 2 ×10−3 × 0. I 0 = I 2 = 0. 2π LC L = 2mH = 2 ×10−3 H . (ii) the rms current through the inductance. e0 = e 2 = 220 × 2 = 311. 50 Hz.sakshieducation.1×10−6 F 7. Sol: 1×105 = 11.m.7sin(120 π t ) V E0 = 169. value 220V and frequency 50Hz. (iii) the peak current and (iv) the peak emf. C = 0.f. ER = 30V .m. At what frequency does maximum current flow? 1 f0 = .f. Find (i) the impedance of the circuit.7 volt = 169. Sol: e. Sol: An inductance of 1 H. capacitance and resistance. Find the emf of the source. X L = Lω = L ( 2π f ) = 1( 2π × 50 ) = 314. Q Z = R 2 + ( X L − X C ) 2 ( 2 ) E = 302 + ( 70 − 30 ) = 302 + 402 = 50V .828π When an inductor. 30 V and 30 V respectively. EL = 70. a capacitor and a resistor are connected in series across a source of alternating emf.2 − 1591) = 1277Ω e 220 = = 0. and its frequency. is e = 169. Sol: iv) The peak emf. e = 220V .2436 A ii) The rms current I = 6. Sol: E 220 = = 20 11A = 66.7 A I0 = 0 = = A= 3. n = 50Hz ω = 2π n = 2π × 50 = 100π rad s −1 R = 1Ω L = 0. c = 2µ F = 2 ×10−6 F 4. ω = 100π rads −1 .14 = 6. of e = 300sin100π t is applied across the terminals of an inductor of inductance 20 mH. and current at resonance/ L = 0.f. of 220V. C = 2µ F = 2×10−6 F For resonance.com V = V0 sin ωt 3. e0 = 200V . current e = 300sin100π t Sol: e0 = 300V .s current.s.14 tan φ = = = ⇒ tan φ = tan 720 ⇒ φ = 720 R 1 1 A 0. Sol: 1 1 5 ×103 5000 = = − = 1591Ω ωc 100π × 2 ×10−6 π π e 200 π Peak current = I 0 = 0 = = A X 0 5000 25 π I0 π I rms = = A = 0.7A Z 11 X π 3. (ii) the peak current and (iii) the r.f.008889A 2 25 2 An alternating sinusoidal e.com .m.414 I 150 ⇒ I rms = = 33.sakshieducation. Sol: X L = ω L = 100 × 20 ×10−3 = 2π = 2 × 3.m.01 = 10−2 H Capacitive reactance = X c = 2 Impedance Z = R 2 + X L = (1) 2 + ( Lω ) 2 = 11Ω 4.03H = 3 ×10−2 H . of e = 200sin100π t is applied across a capacitor of capacitance 2µ F . current and the phase difference between the e. At what frequency will they resonate? What will be the phase difference between the e. 50Hz is connected to circuit of resistance 1Ω and inductance 0. and the current? E = 220V.01H.1sin (100π t ) volt An alternating e.03H inductor.284 Ω 150 e 300 150 = 7. ω = 10π rads −1 . What is the r. a 10Ω resistor and a 2µ F capacitor are connected in series. The peak current and (iii) the r. 1 1 ωL = ⇒ ω2 = LC ωC 1 1 104 ⇒ω= = = rad sec −1 −2 −6 LC 6 (3 × 10 ) (2 ×10 ) Peak current = I 0 = www.f.m.f.m. The capacitive reactance (ii). ⇒ V = 220 2 sin(100π )t = 311.sakshieducation.m. e = 200sin100π t .14 2 π 2 An alternating e. Find (i) the inductive reactance.m.m. R = 10Ω .m.14 X L 2π π 75 ×1. L = 20mH = 20 ×10−3 H 3.f.7 A I rms = 0 = 3.s.www. Find (i). com .d.’s across the inductor. L = 1mH = 10−3 H . It behaves like a pure resistor circuit.28 4. In a series LCR circuit. 3. VR = 40V E = VR2 + (VL − VC ) 2 = 402 + (90 − 60) 2 = 1600 + 900 = 50V ASSESS YOURSELF 1.sakshieducation. Find the e.5 × (2π f ) ×10−3 5. When the emf is maximum how is the current in a pure inductor? Sol: Minimum.d. what is the relation between the potential difference across the inductor and capacitor? www.com ⇒ n= ω 104 = = 649. (ii) the p. across the capacitor. (iv) the p. 5. capacitor. of the source.m. since the phase difference between emf and current through a pure inductor is 900 2. 40V respectively. Sol: 1 105 100 = × 2π × ×10−3 = = 50V 2 2π 2 iv) P. at resonance what is the nature of the circuit? Ans.7 Hz. E = 5V 1 Impedance is minimum when ω L = ωC 1 1 ⇒ ω2 = → (2π f )2 = LC LC 1 ⇒ 4π 2 f 2 = LC 1 1 ⇒ f= = = 1591Hz 2π LC 2π 10−3 ×10−7 Impedance at resonance = Z = R 2 + X 2 = (10) 2 + ( X L − X C ) 2 = 10Ω i) Current at resonance I = ii) P. since the phase difference between and current through a pure capacitor is 900 4.1µ F and R = 10Ω . across the inductor.5 A z 10 Vc = IX c = 0.sakshieducation. Maximum. R = 10Ω .f.f. and resistor are 90V. 60V. across resistor = VR = IR = 0.1µ F = 10−7 F .www. a capacitor and a resistor are connected in series across a source of alternating e. across inductor VL = IX L = I (ω L) = 0.D. Maximum.D. The p.D. (iii) p.d. When the emf is minimum how is the current in a pure capacitor? Ans. In a series LCR circuit. 2π 6 × 6.5 × 10 = 5V An inductor. C = 0. C = 0. Find (i) the frequency at which the impedance is minimum.m. across capacitor E 5 = = 0. across the resistor and (v) the current at resonance. It is connected across a source of alternating emf of 5V but of variable frequency.5 × 1 1 1 = × = wc 2 (2π f ) × 10−7 107 = 50V 105 4π × 2π iii) P. since the phase difference between emf and current through a pure resistor is 00 . When the emf is maximum how is the current in a pure resistor? Ans.d. VL = 90V . VC = 60V . Sol: 1 ωL − X X − Xc ωc tan φ = = L = R R R 1 ⇒ tan φ = 0 ⇒ φ = 00 Since ω L = ωc A series LCR circuit has L = 1 mH. www.sakshieducation.sakshieducation.com . www.com Ans. They are equal and opposite.