Revision test 4
Exercise 52 Further problem on R–C series a.c
circuits
1. A voltage of 35 V is applied across a C–R series circuit. If the voltage across the resistor is 21 V, find the voltage across the capacitor. [28 V]
2. A resistance of 50is connected in series with a capacitance of 20µF. If a supply of 200 V, 100 Hz is connected across the arrangement find (a) the cir- cuit impedance, (b) the current flowing, and (c) the phase angle between voltage and current.
[(a) 93.98(b) 2.128 A (c) 57.86◦leading]
3. An alternating voltage v=250 sin 800t volts is applied across a series circuit containing a 30 resistor and 50µF capacitor. Calculate (a) the cir- cuit impedance, (b) the current flowing, (c) the p.d.
across the resistor, (d) the p.d. across the capaci- tor, and (e) the phase angle between voltage and current.
[(a) 39.05(b) 4.526 A (c) 135.8 V (d) 113.2 V (e) 39.81◦leading]
4. A 400 resistor is connected in series with a 2358 pF capacitor across a 12 V a.c. supply. Deter- mine the supply frequency if the current flowing in
the circuit is 24 mA. [225 kHz]
15.6 R–L–C series a.c. circuit
In an a.c. series circuit containing resistance R, inductance L and capacitance C, the applied voltage V is the phasor sum of VR, VLand VC(see Figure 15.12). VLand VCare anti-phase, i.e. displaced by 180◦, and there are three pha- sor diagrams possible — each depending on the relative values of VLand VC.
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Single-phase series a.c. circuits 175
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When XL>XC(Figure 15.12(b)):
Z=√
[R2+(XL−XC)2] and tanφ=(XL−XC)
R
Figure 15.12
When XC>XL(Figure 15.12(c)):
Z=√
[R2+(XC−XL)2] and tanα=(XC−XL)
R
When XL=XC(Figure 15.12(d)), the applied voltage V and the current I are in phase. This effect is called series resonance (see Section 15.7).
Problem 15. A coil of resistance 5and inductance 120 mH in series with a 100µF capacitor, is connected to a 300 V, 50 Hz supply. Calculate (a) the current flowing, (b) the phase difference between the supply voltage and current, (c) the voltage across the coil and (d) the voltage across the capacitor.
The circuit diagram is shown in Figure 15.13
Figure 15.13
XL=2πfL=2π(50)(120×10−3)=37.70 XC= 1
2πfC = 1
2π(50)(100×10−6)=31.83 Since XLis greater than XCthe circuit is inductive.
XL−XC=37.70−31.83=5.87 Impedance Z=√
[R2+(XL−XC)2]=√
[(5)2+(5.87)2]
=7.71 (a) Current I=V
Z = 300
7.71=38.91 A (b) Phase angleφ=tan−1
XL−XC R
=tan−15.87 5
=49.58◦
=49◦35 (c) Impedance of coil, ZCOIL
=√
(R2+XL2)=√
[(5)2+(37.70)2]=38.03 Voltage across coil VCOIL=IZCOIL=(38.91)(38.03)
=1480 V Phase angle of coil=tan−1XL
R =tan−1 37.70
5
=82.45◦
=82◦27lagging (d) Voltage across capacitor VC=IXC=(38.91)(31.83)
=1239 V The phasor diagram is shown in Figure 15.14. The supply voltage V is the phasor sum of VCOILand VC
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176 Electrical Circuit Theory and Technology
Figure 15.14
Series connected impedances
For series connected impedances the total circuit impedance can be represented as a single L–C–R circuit by combining all values of resistance together, all values of inductance together and all values of capacitance together, (remembering that for series connected capacitors
1 C = 1
C1 + 1
C2 + · · ·).
For example, the circuit of Figure 15.15(a) showing three impedances has an equivalent circuit of Figure 15.15(b).
Figure 15.15
Problem 16. The following three impedances are connected in series across a 40 V, 20 kHz supply: (i) a resistance of 8, (ii) a coil of inductance 130µH and 5resistance, and (iii) a 10resistor in series with a 0.25µF capacitor. Calculate (a) the circuit current, (b) the circuit phase angle and (c) the voltage drop across each impedance.
The circuit diagram is shown in Figure 15.16(a). Since the total circuit resistance is 8+5+10, i.e. 23, an equivalent circuit diagram may be drawn as shown in Figure 15.16(b)
Figure 15.16
Inductive reactance, XL=2πfL
=2π(20×103)(130×10−6)
=16.34 Capacitive reactance, XC= 1
2πfC
= 1
2π(20×103)(0.25×10−6)
=31.83
Since XC>XL, the circuit is capacitive (see phasor diagram in Figure 15.12(c)).
XC−XL=31.83−16.34=15.49. (a) Circuit impedance, Z=√
[R2+(XC−XL)2]
=√
[232+15.492]
=27.73 Circuit current, I=V
Z = 40
27.73=1.442 A
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PART
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(b) From Figure 15.12(c), circuit phase angle φ=tan−1
XC−XL
R
i.e,φ=tan−1 15.49
23
=33.96◦=33◦58leading (c) From Figure 15.16(a), V1=IR1=(1.442)(8)
=11.54 V V2=IZ2=I√
(52+16.342)=(1.442)(17.09)
=24.64 V V3=IZ3=I√
(102+31.832)=(1.442)(33.36)
=48.11 V
The 40 V supply voltage is the phasor sum of V1, V2
and V3
Problem 17. Determine the p.d.’s V1 and V2 for the circuit shown in Figure 15.17 if the frequency of the supply is 5 kHz. Draw the phasor diagram and hence determine the supply voltage V and the circuit phase angle.
Figure 15.17 For impedance Z1:
R1 = 4and XL = 2πfL=2π(5×103)(0.286×10−3)
=8.985 V1= IZ1=I√
(R2+XL2)=5√
(42+8.9852)=49.18 V Phase angleφ1=tan−1
XL
R
=tan−1 8.985
4
=66◦0lagging For impedance Z2:
R2=8and XC = 1
2πfC = 1
2π(5×103)(1.273×10−6)
=25.0 V2 = IZ2= I√
(R2+XC2)=5√
(82+25.02)=131.2 V Phase angleφ2=tan−1
XC R
=tan−1 25.0
8
=72◦15leading
The phasor diagram is shown in Figure 15.18.
Figure 15.18
The phasor sum of V1and V2gives the supply voltage V of 100 V at a phase angle of 53◦8leading. These values may be determined by drawing or by calculation — either by resolving into horizontal and vertical components or by the cosine and sine rules.
Now try the following exercise.
Exercise 53 Further problems on R–L–C series a.c. circuits
1. A 40µF capacitor in series with a coil of resistance 8and inductance 80 mH is connected to a 200 V, 100 Hz supply. Calculate (a) the circuit impedance, (b) the current flowing, (c) the phase angle between voltage and current, (d) the voltage across the coil, and (e) the voltage across the capacitor.
[(a) 13.18(b) 15.17 A (c) 52.63◦lagging (d) 772.1 V (e) 603.6 V]
2. Find the values of resistance R and inductance L in the circuit of Figure 15.19.
[R=131, L=0.545 H]
240 V, 50 Hz I = 1.5∠−35° A
R L 40 µF
Figure 15.19
3. Three impedances are connected in series across a 100 V, 2 kHz supply. The impedances comprise:
(i) an inductance of 0.45 mH and 2resistance, (ii) an inductance of 570µH and 5resistance,
and
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178 Electrical Circuit Theory and Technology
(iii) a capacitor of capacitance 10µF and resis- tance 3.
Assuming no mutual inductive effects between the two inductances calculate (a) the circuit impedance, (b) the circuit current, (c) the circuit phase angle and (d) the voltage across each impedance.
[(a) 11.12(b) 8.99 A (c) 25.92◦lagging (d) 53.92 V, 78.53 V, 76.46 V]
4. For the circuit shown in Figure 15.20 determine the voltages V1and V2if the supply frequency is 1 kHz.
Draw the phasor diagram and hence determine the supply voltage V and the circuit phase angle.
[V1=26.0 V, V2=67.05 V, V=50 V, 53.14◦leading]
Figure 15.20
15.7 Series resonance
As stated in Section 15.6, for an R–L–C series circuit, when XL=XC (Figure 15.12(d)), the applied voltage V and the current I are in phase. This effect is called series resonance. At resonance:
(i) VL=VC
(ii) Z=R (i.e. the minimum circuit impedance possible in an L–C–R circuit)
(iii) I=V
R (i.e. the maximum current possible in an L–C–R circuit)
(iv) Since XL=XC, then 2πfrL= 1 2πfrC from which, fr2= 1
(2π)2LC
and, fr= 1
2π√ (LC)Hz
where fris the resonant frequency.
(v) The series resonant circuit is often described as an acceptor circuit since it has its minimum impedance, and thus maximum current, at the resonant frequency.
(vi) Typical graphs of current I and impedance Z against frequency are shown in Figure 15.21.
Figure 15.21
Problem 18. A coil having a resistance of 10and an inductance of 125 mH is connected in series with a 60µF capacitor across a 120 V supply. At what fre- quency does resonance occur? Find the current flowing at the resonant frequency.
Resonant frequency, fr= 1 2π√
(LC)Hz
= 1
2π
125 103
60 106
Hz
= 1
2π
125×6 108
= 1
2π
√[(125)(6)]
104
= 104 2π√
[(125)(6)] =58.12 Hz At resonance, XL=XCand impedance Z=R Hence current, I=V
R=120 10 =12 A
Problem 19. The current at resonance in a series L–
C–R circuit is 100µA. If the applied voltage is 2 mV at a frequency of 200 kHz, and the circuit inductance is 50µH, find (a) the circuit resistance, and (b) the circuit capacitance.
(a) I=100µA=100×10−6A; V=2 mV=2×10−3V At resonance, impedance Z=resistance R
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Hence R= V
I = 20×10−3
100×10−6 = 2×106
100×103 =20 (b) At resonance XL=XC
i.e. 2πfL= 1 2πfC Hence capacitance
C= 1
(2πf )2L
= 1
(2π×200×103)2(50×10−6)F
= (106)(106) (4π)2(1010)(50)µF
=0.0127µF or 12.7 nF
15.8 Q-factor
At resonance, if R is small compared with XLand XC, it is possible for VL and VC to have voltages many times greater than the supply voltage (see Figure 15.12(d)).
Voltage magnification at resonance
= voltage across L (or C) supply voltage V
This ratio is a measure of the quality of a circuit (as a resonator or tuning device) and is called the Q-factor.
Hence Q-factor=VL V =IXL
IR =XL
R =2πfrL R Alternatively, Q-factor=VC
V =IXC IR =XC
R = 1
2πfrCR At resonance fr= 1
2π√
(LC)i.e. 2πfr= 1
√(LC) Hence Q-factor=2πfrL
R = 1
√(LC) L
R
= 1 R
L C
(Q-factor is explained more fully in Chapter 28, page 349) Problem 20. A coil of inductance 80 mH and negli- gible resistance is connected in series with a capac- itance of 0.25µF and a resistor of resistance 12.5 across a 100 V, variable frequency supply. Determine (a) the resonant frequency, and (b) the current at reso- nance. How many times greater than the supply voltage is the voltage across the reactances at resonance?
(a) Resonant frequency fr
= 1
2π 80
103
0.25 106
= 1 2π
(8)(0.25) 108
= 104 2π√ 2
=1125.4 Hz=1.1254 kHz (b) Current at resonance I=V
R= 100 12.5=8 A Voltage across inductance, at resonance, VL=IXL=(I)(2πfL)
=(8)(2π)(1125.4)(80×10−3)
=4525.5 V
(Also, voltage across capacitor, VC=IXC= I
2πfC = 8
2π(1125.4)(0.25×10−6)
=4525.5 V ) Voltage magnification at resonance=VL
V orVc
V
=4525.5 100
=45.255 V i.e. at resonance, the voltage across the reactances are 45.255 times greater than the supply voltage. Hence Q-factor of circuit is 45.255.
Problem 21. A series circuit comprises a coil of resistance 2 and inductance 60 mH, and a 30µF capacitor. Determine the Q-factor of the circuit at resonance.
At resonance, Q-factor= 1 R
L C
=1 2
60×10−3 30×10−6
=1 2
60×106 30×103
=1 2
√(2000)
=22.36
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180 Electrical Circuit Theory and Technology
Problem 22. A coil of negligible resistance and inductance 100 mH is connected in series with a capacitance of 2µF and a resistance of 10across a 50 V, variable frequency supply. Determine (a) the res- onant frequency, (b) the current at resonance, (c) the voltages across the coil and the capacitor at resonance, and (d) the Q-factor of the circuit.
(a) Resonant frequency, fr= 1 2π√
(LC)
= 1
2π
100 103
2 106
= 1
2π 20
108
= 1 2π√
20 104
= 104 2π√ 20
=355.9 Hz (b) Current at resonance I=V
R=50 10=5 A (c) Voltage across coil at resonance,
VL=IXL=I(2πfrL)
=(5)(2π×355.9×100×10−3)
=1118 V
Voltage across capacitance at resonance, VC=IXC= I
2πfrC
= 5
2π(355.9)(2×10−6)
=1118 V
(d) Q-factor (i.e. voltage magnification at resonance)
=VL
V orVC
V
=1118
50 =22.36
Q-factor may also have been determined by2πfrL R or 1
2πfrCR or 1 R
L C
Now try the following exercise.
Exercise 54 Further problems on series resonance