At the end of this chapter you should be able to:
• calculate unknown voltages, currents and resis- tances in a series circuit
• understand voltage division in a series circuit
• calculate unknown voltages, currents and resist- ances in a parallel network
• calculate unknown voltages, currents and resist- ances in series-parallel networks
• understand current division in a two-branch parallel network
• understand and perform calculations on relative and absolute voltages
• describe the advantages and disadvantages of series and parallel connection of lamps
5.1 Series circuits
Figure 5.1 shows three resistors R1, R2and R3connected end to end, i.e. in series, with a battery source of V volts.
Since the circuit is closed a current I will flow and the p.d.
across each resistor may be determined from the voltmeter readings V1, V2and V3
Figure 5.1 In a series circuit
(a) the current I is the same in all parts of the circuit and hence the same reading is found on each of the two ammeters shown, and
(b) the sum of the voltages V1, V2and V3is equal to the total applied voltage, V , i.e.
V=V1+V2+V3
From Ohm’s law:
V1=IR1, V2=IR2, V3=IR3and V=IR where R is the total circuit resistance.
Since V=V1+V2+V3
then IR =IR1+IR2+IR3
Dividing throughout by I gives R=R1+R2+R3
Thus for a series circuit, the total resistance is obtained by adding together the values of the separate resistances.
Problem 1. For the circuit shown in Figure 5.2, deter- mine (a) the battery voltage V , (b) the total resistance of the circuit, and (c) the values of resistance of resis- tors R1, R2and R3, given that the p.d.’s across R1, R2 and R3are 5 V, 2 V and 6 V respectively.
Figure 5.2
(a) Battery voltage V=V1+V2+V3
=5+2+6=13 V (b) Total circuit resistance R=V
I =13
4 =3.25 (c) Resistance R1=V1
I =5
4=1.25 Resistance R2=V2
I =2 4=0.5 Resistance R3=V3
I =6 4=1.5
(Check: R1+R2+R3=1.25+0.5+1.5=3.25=R)
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Problem 2. For the circuit shown in Figure 5.3, deter- mine the p.d. across resistor R3. If the total resistance of the circuit is 100, determine the current flowing through resistor R1. Find also the value of resistor R2.
Figure 5.3
P.d. across R3, V3=25−10−4=11 V Current I=V
R= 25
100=0.25 A, which is the current flowing in each resistor
Resistance R2=V2 I = 4
0.25=16
Problem 3. A 12 V battery is connected in a circuit having three series-connected resistors having resist- ances of 4, 9 and 11. Determine the current flowing through, and the p.d. across the 9 resistor.
Find also the power dissipated in the 11resistor.
The circuit diagram is shown in Figure 5.4.
Figure 5.4
Total resistance R=4+9+11=24 Current I=V
R=12
24=0.5 A, which is the current in the 9resistor.
P.d. across the 9resistor, V1=I×9=0.5×9
=4.5 V
Power dissipated in the 11resistor, P=I2R=0.52(11)
=0.25(11)
=2.75 W
5.2 Potential divider
The voltage distribution for the circuit shown in Figure 5.5(a) is given by:
V1= R1
R1+R2
V
V2= R2
R1+R1
V
Figure 5.5
The circuit shown in Figure 5.5(b) is often referred to as a potential divider circuit. Such a circuit can consist of a number of similar elements in series connected across a voltage source, voltages being taken from connections between the elements. Frequently the divider consists of two resistors as shown in Figure 5.5(b), where
VOUT= R2
R1+R2
VIN
A potential divider is the simplest way of producing a source of lower e.m.f. from a source of higher e.m.f., and is the basic operating mechanism of the potentiometer, a measuring device for accurately measuring potential differences (see page 99).
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30 Electrical Circuit Theory and Technology
Problem 4. Determine the value of voltage V shown in Figure 5.6.
Figure 5.6
Figure 5.6 may be redrawn as shown in Figure 5.7, and voltage
V= 6
6+4
(50)=30 V
Figure 5.7
Problem 5. Two resistors are connected in series across a 24 V supply and a current of 3 A flows in the circuit. If one of the resistors has a resistance of 2 determine (a) the value of the other resistor, and (b) the p.d. across the 2resistor. If the circuit is connected for 50 hours, how much energy is used?
The circuit diagram is shown in Figure 5.8
Figure 5.8
(a) Total circuit resistance R=V I =24
3 =8 Value of unknown resistance, Rx=8−2=6 (b) P.d. across 2resistor, V1=IR1=3×2=6 V
Alternatively, from above, V1=
R1 R1+Rx
V=
2 2+6
(24)=6 V
Energy used=power×time
=V×I×t
=(24×3 W) (50 h)
=3600 Wh=3.6 kWh Now try the following exercise.
Exercise 10 Further problems on series circuits 1. The p.d’s measured across three resistors connected
in series are 5 V, 7 V and 10 V, and the supply current is 2 A. Determine (a) the supply voltage, (b) the total circuit resistance and (c) the values of the three resistors.
[(a) 22 V (b) 11(c) 2.5, 3.5, 5] 2. For the circuit shown in Figure 5.9, determine the value of V1. If the total circuit resistance is 36 determine the supply current and the value of resistors R1, R2and R3.
[10 V, 0.5 A, 20, 10, 6]
Figure 5.9
3. When the switch in the circuit in Figure 5.10 is closed the reading on voltmeter 1 is 30 V and that on voltmeter 2 is 10 V. Determine the reading on the ammeter and the value of resistor Rx.
[4 A, 2.5]
Figure 5.10
4. Two resistors are connected in series across an 18 V supply and a current of 5 A flows. If one of the resistors has a value of 2.4 determine (a) the value of the other resistor and (b) the p.d. across the 2.4resistor. [(a) 1.2(b) 12 V]
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5. An arc lamp takes 9.6 A at 55 V. It is operated from a 120 V supply. Find the value of the stabilizing resistor to be connected in series. [6.77] 6. An oven takes 15 A at 240 V. It is required to reduce the current to 12 A. Find (a) the resistor which must be connected in series, and (b) the voltage across
the resistor. [(a) 4(b) 48 V]
5.3 Parallel networks
Figure 5.11 shows three resistors, R1, R2and R3connected across each other, i.e. in parallel, across a battery source of V volts.
Figure 5.11
In a parallel circuit:
(a) the sum of the currents I1, I2 and I3 is equal to the total circuit current, I, i.e. I=I1+I2+I3, and (b) the source p.d., V volts, is the same across each of the
resistors.
From Ohm’s law:
I1 = V
R1, I2= V
R2, I3= V
R3 and I= V R
where R is the total circuit resistance.
Since I=I1+I2+I3
then V R= V
R1 + V R2 + V
R3
Dividing throughout by V gives:
1 R= 1
R1+ 1 R2+ 1
R3
This equation must be used when finding the total resist- ance R of a parallel circuit. For the special case of two resistors in parallel
1 R= 1
R1 + 1
R2 = R2+R1 R1R2
Hence R= R1R2
R1+R2
i.e. product sum
Problem 6. For the circuit shown in Figure 5.12, determine (a) the reading on the ammeter, and (b) the value of resistor R2.
Figure 5.12
P.d. across R1is the same as the supply voltage V . Hence supply voltage, V=8×5=40 V
(a) Reading on ammeter, I= V R3 =40
20=2 A (b) Current flowing through R2=11−8−2=1 A
Hence, R2=V I2=40
1 =40
Problem 7. Two resistors, of resistance 3and 6, are connected in parallel across a battery having a volt- age of 12 V. Determine (a) the total circuit resistance and (b) the current flowing in the 3resistor.
Ch05-H8139.tex 2/4/2007 9: 26 page 32
32 Electrical Circuit Theory and Technology The circuit diagram is shown in Figure 5.13.
Figure 5.13
(a) The total circuit resistance R is given by 1
R = 1 R1 + 1
R2 = 1 3+1
6 1
R = 2+1
6 = 3
6 Hence, R=6
3 =2
Alternatively, R= R1R2
R1+R2 =3×6 3+6=18
9 =2
(b) Current in the 3resistance, I1 = V R1 = 12
3 =4 A Problem 8. For the circuit shown in Figure 5.14, find (a) the value of the supply voltage V and (b) the value of current I.
Figure 5.14
(a) P.d. across 20resistor=I2R2 =3×20=60 V, hence supply voltage V=60 V since the circuit is connected in parallel.
(b) Current I1= V R1 = 60
10 =6 A; I2=3 A I3= V
R3 = 60 60 =1 A
Current I=I1+I2+I3and hence I=6+3+1=10 A
Alternatively,1 R= 1
60+ 1 20+ 1
10 = 1+3+6
60 =10
60 Hence total resistance R=60
10=6 Current I=V
R=60 6 =10 A
Problem 9. Given four 1resistors, state how they must be connected to give an overall resistance of (a) 14(b) 1(c) 113(d) 212, all four resistors being connected in each case.
(a) All four in parallel (see Figure 5.15), since 1
R=1 1+1
1+1 1+1
1 = 4
1, i.e. R=1 4
Figure 5.15
(b) Two in series, in parallel with another two in series (see Figure 5.16), since 1and 1in series gives 2, and 2in parallel with 2gives: 2×2
2+2=4 4=1
Figure 5.16
(c) Three in parallel, in series with one (see Fig- ure 5.17), since for the three in parallel,
1 R=1
1+1 1+1
1 = 3
1, i.e. R=1
3and1
3in series with 1gives 113
Figure 5.17
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Series and parallel networks 33
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(d) Two in parallel, in series with two in series (see Figure 5.18), since for the two in parallel
R=1×1 1+1=1
2, and 1
2, 1and 1in series gives 21
2
Figure 5.18
Problem 10. Find the equivalent resistance for the circuit shown in Figure 5.19.
Figure 5.19
R3, R4and R5are connected in parallel and their equivalent resistance R is given by:
1 R= 1
3+1 6+ 1
18 = 6+3+1
18 =10
18 Hence R=18
10=1.8
The circuit is now equivalent to four resistors in series and the equivalent circuit resistance=1+2.2+1.8+4=9
5.4 Current division
For the circuit shown in Figure 5.20, the total circuit resistance, RT is given by:
RT = R1R2 R1+R2
Figure 5.20
and V=IRT=I
R1R2 R1+R2
Current I1= V R1= I
R1
R1R2
R1+R2
= R2
R1+R2
(I) Similarly,
current I2= V R2= I
R2
R1R2 R1+R2
= R1
R1+R2
(I) Summarizing, with reference to Figure 5.20
I1= R2
R1+R2
(I) and I2= R1
R1+R2
(I)
Problem 11. For the series-parallel arrangement shown in Figure 5.21, find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.
Figure 5.21
(a) The equivalent resistance Rxof R2 and R3 in paral- lel is:
Rx= 6×2 6+2 = 12
8 =1.5
The equivalent resistance RT of R1, Rx and R4 in series is:
RT =2.5+1.5+4=8 Supply current I= V
RT = 200 8 =25 A (b) The current flowing through R1and R4is 25 A
The current flowing through R2
= R3
R2+R3
I=
2 6+2
25
=6.25 A
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34 Electrical Circuit Theory and Technology The current flowing through R3
= R2
R2+R3
I= 6
6+2
25
=18.75 A
(Note that the currents flowing through R2 and R3
must add up to the total current flowing into the parallel arrangement, i.e. 25 A)
(c) The equivalent circuit of Figure 5.21 is shown in Figure 5.22.
p.d. across R1, i.e. V1=IR1=(25)(2.5)=62.5 V p.d. across Rx, i.e. Vx=IRx=(25)(1.5)=37.5 V p.d. across R4, i.e. V4=IR4=(25)(4)=100 V Hence the p.d. across R2=p.d. across R3=37.5 V
Figure 5.22
Problem 12. For the circuit shown in Figure 5.23 cal- culate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and (b) the current flowing in each of the four resistors.
Figure 5.23
(a) Power dissipated P=VI watts, hence 2500=(250)(I) i.e.I= 2500
250 =10 A From Ohm’s law, RT=V
I =250
10 =25, where RT is the equivalent circuit resistance.
The equivalent resistance of R1and R2in parallel is 15×10
15+10 = 150 25 =6
The equivalent resistance of resistors R3 and Rx in parallel is equal to 25−6, i.e. 19.
There are three methods whereby Rx can be deter- mined.
Method 1
The voltage V1=IR, where R is 6, from above, i.e. V1=(10)(6)=60 V
Hence V2=250 V−60 V=190 V=p.d. across R3
=p.d. across Rx
I3= V2 R3=190
38 =5 A.Thus I4=5 A also, since I=10 A
Thus Rx=V2 I4 =190
5 =38 Method 2
Since the equivalent resistance of R3 and Rx in parallel is 19,
19= 38Rx
38+Rx
i.e. product sum
Hence 19(38+Rx)=38Rx 722+19Rx=38Rx
722=38Rx−19Rx=19Rx Thus Rx=722
19 =38 Method 3
When two resistors having the same value are con- nected in parallel the equivalent resistance is always half the value of one of the resistors. Thus, in this case, since RT=19 and R3=38, then Rx=38could have been deduced on sight.
(b) Current I1= R2
R1+R2
I=
10 15+10
(10)
= 2
5
(10)=4 A Current I2=
R1
R1+R2
I=
15 15+10
(10)
= 3
5
(10)=6 A From part (a), method 1, I3=I4=5 A
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Problem 13. For the arrangement shown in Fig- ure 5.24, find the current Ix.
Figure 5.24
Commencing at the right-hand side of the arrangement shown in Figure 5.24, the circuit is gradually reduced in stages as shown in Figure 5.25(a)–(d).
Figure 5.25
From Figure 5.25(d), I= 17 4.25=4 A From Figure 5.25(b), I1=
9 9+3
(I)=
9 12
(4)=3 A
From Figure 5.24, Ix= 2
2+8
(I1)= 2
10
(3)=0.6 A
Now try the following exercise.
Exercise 11 Further problems on parallel networks
1. Resistances of 4and 12are connected in paral- lel across a 9 V battery. Determine (a) the equivalent circuit resistance, (b) the supply current, and (c) the current in each resistor.
[(a) 3(b) 3 A (c) 2.25 A, 0.75 A]
2. For the circuit shown in Figure 5.26 determine (a) the reading on the ammeter, and (b) the value
of resistor R. [2.5 A, 2.5]
Figure 5.26
3. Find the equivalent resistance when the following resistances are connected (a) in series, (b) in parallel (i) 3and 2 (ii) 20 kand 40 k (iii) 4,8and 16(iv) 800,4 kand 1500
[(a) (i) 5(ii) 60 k(iii) 28(iv) 6.3 k (b) (i) 1.2(ii) 13.33 k(iii) 2.29(iv) 461.5] 4. Find the total resistance between terminals A and B of the circuit shown in Figure 5.27(a) [8]
Figure 5.27
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36 Electrical Circuit Theory and Technology
5. Find the equivalent resistance between terminals C and D of the circuit shown in Figure 5.27(b)
[27.5] 6. Resistors of 20, 20and 30are connected in parallel. What resistance must be added in series with the combination to obtain a total resistance of 10. If the complete circuit expends a power of 0.36 kW, find the total current flowing.[2.5, 6 A]
7. (a) Calculate the current flowing in the 30 resistor shown in Figure 5.28.
Figure 5.28
(b) What additional value of resistance would have to be placed in parallel with the 20and 30 resistors to change the supply current to 8 A, the supply voltage remaining constant.
[(a) 1.6 A (b) 6] 8. Determine the currents and voltages indicated in
the circuit shown in Figure 5.29.
[I1=5 A, I2=2.5 A, I3=123A, I4=56A I5 =3 A, I6=2 A, V1=20 V, V2=5 V, V3=6 V]
Figure 5.29
9. Find the current I in Figure 5.30. [1.8 A]
Figure 5.30
10. A resistor of 2.4 is connected in series with another of 3.2 . What resistance must be placed across the one of 2.4so that the total resistance of the circuit shall be 5? [7.2] 11. A resistor of 8is connected in parallel with one of 12and the combination is connected in series with one of 4. A p.d. of 10 V is applied to the circuit. The 8resistor is now placed across the 4resistor. Find the p.d. required to send the same current through the 8resistor. [30 V]
5.5 Relative and absolute voltages
In an electrical circuit, the voltage at any point can be quoted as being ‘with reference to’ (w.r.t.) any other point in the circuit. Consider the circuit shown in Figure 5.31.
The total resistance,
RT=30+50+5+15=100 and current, I= 200
100 =2 A
I = 2 A 200 V
30Ω 50Ω
5Ω
15Ω
A B
C Figure 5.31
If a voltage at point A is quoted with reference to point B then the voltage is written as VAB. This is known as a ‘relative voltage’. In the circuit shown in Figure 5.31, the voltage at A w.r.t. B is I×50, i.e.
2×50=100 V and is written as VAB=100 V.
It must also be indicated whether the voltage at A w.r.t.
B is closer to the positive terminal or the negative terminal of the supply source. Point A is nearer to the positive terminal than B so is written as VAB=100 V or VAB= +100 V or VAB= 100 V+ve.
If no positive or negative is included, then the voltage is always taken to be positive.
If the voltage at B w.r.t. A is required, then VBAis neg- ative and written as VBA= −100 V or VBA=100 V−ve.
If the reference point is changed to the earth point then any voltage taken w.r.t. the earth is known as an ‘absolute potential’. If the absolute voltage of A in Figure 5.31 is required, then this will be the sum of the voltages across the 50and 5resistors, i.e. 100+10=110 V and is written as VA=110 V or VA= +110 V or VA=110 V
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Series and parallel networks 37
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+ve, positive since moving from the earth point to point A is moving towards the positive terminal of the source. If the voltage is negative w.r.t. earth then this must be indi- cated; for example, VC=30 V negative w.r.t. earth, and is written as VC= −30 V or VC=30 V−ve.
Problem 14. For the circuit shown in Fig. 5.32, cal- culate (a) the voltage drop across the 4 k resistor, (b) the current through the 5 kresistor, (c) the power developed in the 1.5 kresistor, (d) the voltage at point X w.r.t. earth, and (e) the absolute voltage at point X.
X 1 kΩ 4 kΩ
5 kΩ
1.5 kΩ 24 V
Figure 5.32
(a) Total circuit resistance, RT=[(1+4)kin parallel with 5 k] in series with 1.5 k
i.e. RT= 5×5
5+5+1.5=4 k Total circuit current, IT= V
RT= 24
4×103 =6 mA By current division, current in top branch
= 5
5+1+4
×6=3 mA
Hence, volt drop across 4 kresistor
=3×10−3×4×103=12 V (b) Current through the 5 kresistor
=
1+4 5+1+4
×6=3 mA (c) Power in the 1.5 kresistor
=I2TR=(6×10−3)2(1.5×103)=54 mW
(d) The voltage at the earth point is 0 volts. The volt drop across the 4 kis 12 V, from part (a). Since moving from the earth point to point X is moving towards the negative terminal of the voltage source, the voltage at point X w.r.t. earth is−12 V
(e) The ‘absolute voltage at point X’ means the ‘voltage at point X w.r.t. earth’, hence the absolute voltage at point X is−12 V. Questions (d) and (e) mean the same thing.
Now try the following exercise.
Exercise 12 Further problems on relative and absolute voltages
1. For the circuit of Figure 5.33, calculate (a) the abso- lute voltage at points A, B and C, (b) the voltage at A relative to B and C, and (c) the voltage at D relative to B and A.
[(a)+40 V,+29.6 V,+24 V (b)+10.4 V,+16 V (c)−5.6 V,−16 V]
A B
C D
100 V
15Ω 13Ω
5Ω
6Ω
7Ω
Figure 5.33
2. For the circuit shown in Figure 5.34, calculate (a) the voltage drop across the 7 resistor, (b) the current through the 30 resistor, (c) the power developed in the 8 resistor, (d) the voltage at point X w.r.t. earth, and (e) the absolute voltage at point X.
[(a) 1.68 V (b) 0.16 A (c) 460.8 mW (d)+2.88 V (e)+2.88 V]
8Ω 18Ω
7Ω
30Ω
12 V
5Ω X
Figure 5.34
3. In the bridge circuit of Figure 5.35 calculate (a) the absolute voltages at points A and B, and (b) the voltage at A relative to B. [(a) 10 V, 10 V (b) 0 V]
2 kΩ
B A
30 V 1 kΩ
16Ω 8Ω
Figure 5.35
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38 Electrical Circuit Theory and Technology 5.6 Wiring lamps in series and in parallel Series connection
Figure 5.36 shows three lamps, each rated at 240 V, connected in series across a 240 V supply.
Figure 5.36
(i) Each lamp has only240
3 V, i.e. 80 V across it and thus each lamp glows dimly.
(ii) If another lamp of similar rating is added in series with the other three lamps then each lamp now has
240
4 V, i.e. 60 V across it and each now glows even more dimly.
(iii) If a lamp is removed from the circuit or if a lamp develops a fault (i.e. an open circuit) or if the switch is opened then the circuit is broken, no current flows, and the remaining lamps will not light up.
(iv) Less cable is required for a series connection than for a parallel one.
The series connection of lamps is usually limited to decorative lighting such as for Christmas tree lights.
Parallel connection
Figure 5.37 shows three similar lamps, each rated at 240 V, connected in parallel across a 240 V supply.
Figure 5.37
(i) Each lamp has 240 V across it and thus each will glow brilliantly at their rated voltage.
(ii) If any lamp is removed from the circuit or develops a fault (open circuit) or a switch is opened, the remaining lamps are unaffected.
(iii) The addition of further similar lamps in parallel does not affect the brightness of the other lamps.
(iv) More cable is required for parallel connection than for a series one.
The parallel connection of lamps is the most widely used in electrical installations.
Problem 15. If three identical lamps are connected in parallel and the combined resistance is 150, find the resistance of one lamp.
Let the resistance of one lamp be R, then, 1
150=1 R+1
R+1 R=3
R, from which, R=3×150
=450
Problem 16. Three identical lamps A, B and C are connected in series across a 150 V supply. State (a) the voltage across each lamp, and (b) the effect of lamp C failing.
(a) Since each lamp is identical and they are connected in series there is150
3 V, i.e. 50 V across each.
(b) If lamp C fails, i.e. open circuits, no current will flow and lamps A and B will not operate.
Now try the following exercise.
Exercise 13 Further problems on wiring lamps in series and parallel
1. If four identical lamps are connected in parallel and the combined resistance is 100, find the
resistance of one lamp. [400]
2. Three identical filament lamps are connected (a) in series, (b) in parallel across a 210 V supply. State for each connection the p.d. across each lamp.
[(a) 70 V (b) 210 V]
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