13 D.c. circuit theory
Exercise 41 Further problems on the superposi- tion theorem
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140 Electrical Circuit Theory and Technology 6. (a) Resultant current in the 18resistor
=I3−I6
=0.167−0.094
=0.073 A
P.d. across the 18resistor
=0.073×18=1.314 V (b) Resultant current in the 8 V battery
=I1+I5=1.667+0.562
=2.229 A (discharging) (c) Resultant current in the 3 V battery
=I2+I4=1.500+0.656
=2.156 A (discharging) Now try the following exercise.
Exercise 41 Further problems on the superposi-
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D.c. circuit theory 141
PART
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(iii) For the circuit shown in Figure 13.28(a) representing a practical source supplying energy, V=E−Ir, where E is the battery e.m.f., V is the battery terminal voltage and r is the internal resistance of the battery (as shown in Section 4.5). For the circuit shown in Figure 13.28(b), V=E−(−I)r, i.e. V=E+Ir
Figure 13.28
(iv) The resistance ‘looking-in’ at terminals AB in Fig- ure 13.29(a) is obtained by reducing the circuit in stages as shown in Figures 13.29(b) to (d). Hence the equivalent resistance across AB is 7
Figure 13.29
(v) For the circuit shown in Figure 13.30(a), the 3 resistor carries no current and the p.d. across the 20resistor is 10 V. Redrawing the circuit gives Figure 13.30(b), from which
E= 4
4+6
×10=4 V
(vi) If the 10 V battery in Figure 13.30(a) is removed and replaced by a short-circuit, as shown in
Figure 13.30
Figure 13.30(c), then the 20 resistor may be removed. The reason for this is that a short- circuit has zero resistance, and 20 in parallel with zero ohms gives an equivalent resistance of:
(20×0/20+0), i.e. 0. The circuit is then as shown in Figure 13.30(d), which is redrawn in Fig- ure 13.30(e). From Figure 13.30(e), the equivalent resistance across AB,
r=6×4
6+4+3=2.4+3=5.4
(vii) To find the voltage across AB in Figure 13.31:
Since the 20 V supply is across the 5and 15 resistors in series then, by voltage division, the voltage drop across AC,
Figure 13.31 VAC =
5 5+15
(20)=5 V
Similarly, VCB= 12
12+3
(20)=16 V.
VCis at a potential of+20 V.
VA=VC−VAC= +20−5=15 V and VB=VC−VBC= +20−16=4 V.
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142 Electrical Circuit Theory and Technology
Hence the voltage between AB is VA−VB= 15−4=11 V and current would flow from A to B since A has a higher potential than B.
(viii) In Figure 13.32(a), to find the equivalent resistance across AB the circuit may be redrawn as in Fig- ures 13.32(b) and (c). From Figure 13.32(c), the equivalent resistance across AB
= 5×15
5+15+12×3
12+3=3.75+2.4=6.15
Figure 13.32
(ix) In the worked problems in Sections 13.5 and 13.7 following, it may be considered that Thévenin’s and Norton’s theorems have no obvious advantages compared with, say, Kirchhoff’s laws. However, these theorems can be used to analyse part of a cir- cuit and in much more complicated networks the principle of replacing the supply by a constant volt- age source in series with a resistance (or impedance) is very useful.
13.5 Thévenin’s theorem Thévenin’s theorem states:
‘The current in any branch of a network is that which would result if an e.m.f. equal to the p.d. across a break made in the branch, were introduced into the branch, all other e.m.f.’s being removed and represented by the internal resistances of the sources.’
The procedure adopted when using Thévenin’s theorem is summarized below. To determine the current in any branch of an active network (i.e. one containing a source of e.m.f.):
(i) remove the resistance R from that branch,
(ii) determine the open-circuit voltage, E, across the break,
(iii) remove each source of e.m.f. and replace them by their internal resistances and then determine the resistance, r, ‘looking-in’ at the break,
(iv) determine the value of the current from the equivalent circuit shown in Figure 13.33, i.e. I= E
R+r
Figure 13.33
Problem 7. Use Thévenin’s theorem to find the cur- rent flowing in the 10resistor for the circuit shown in Figure 13.34(a).
Following the above procedure:
(i) The 10resistance is removed from the circuit as shown in Figure 13.34(b)
(ii) There is no current flowing in the 5resistor and current I1is given by:
I1= 10
R1+R2 = 10 2+8 =1 A
Figure 13.34
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P.d. across R2=I1R2=1×8=8 V
Hence p.d. across AB, i.e. the open-circuit voltage across the break, E=8 V
(iii) Removing the source of e.m.f. gives the circuit of Figure 13.34(c).
Resistance, r=R3+ R1R2
R1+R2=5+2×8 2+8
=5+1.6=6.6
(iv) The equivalent Thévenin’s circuit is shown in Fig- ure 13.34(d).
Current I= E
R+r= 8
10+6.6= 8
16.6=0.482 A Hence the current flowing in the 10 resistor of Figure 13.34(a) is 0.482 A
Problem 8. For the network shown in Figure 13.35(a) determine the current in the 0.8 resistor using Thévenin’s theorem.
Following the procedure:
(i) The 0.8 resistor is removed from the circuit as shown in Figure 13.35(b).
(ii) Current I1 = 12
1+5+4=12 10=1.2 A
P.d. across 4resistor=4I1=(4) (1.2)=4.8 V Hence p.d. across AB, i.e. the open-circuit voltage across AB, E=4.8 V
Figure 13.35
(iii) Removing the source of e.m.f. gives the circuit shown in Figure 13.35(c). The equivalent circuit of Fig- ure 13.35(c) is shown in Figure 13.35(d), from which, resistance r=4×6
4+6=24 10=2.4
(iv) The equivalent Thévenin’s circuit is shown in Figure 13.35(e), from which,
current I = E
r+R= 4.8
2.4+0.8=4.8 3.2
I=1.5A=current in the 0.8resistor Problem 9. Use Thévenin’s theorem to determine the current I flowing in the 4 resistor shown in Fig- ure 13.36(a). Find also the power dissipated in the 4 resistor.
Following the procedure:
(i) The 4resistor is removed from the circuit as shown in Figure 13.36(b).
Figure 13.36
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144 Electrical Circuit Theory and Technology
(ii) Current I1=E1−E2
r1+r2 =4−2 2+1=2
3A P.d. across AB, E=E1−I1r1=4−2
3
(2)=223V (see Section 13.4(iii))
(Alternatively, p.d. across AB, E=E2−I1r2
=2−2
3
(1)=223V
(iii) Removing the sources of e.m.f. gives the circuit shown in Figure 13.36(c), from which resistance r= 2×1
2+1= 2 3
(iv) The equivalent Thévenin’s circuit is shown in Fig- ure 13.36(d), from which,
current, I= E
r+R= 223
2
3+4 = 8/3 14/3
= 8
14 =0.571 A
=current in the 4resistor Power dissipated in 4resistor,
P=I2R=(0.571)2(4)=1.304 W
Problem 10. Use Thévenin’s theorem to determine the current flowing in the 3resistance of the net- work shown in Figure 13.37(a). The voltage source has negligible internal resistance.
(Note the symbol for an ideal voltage source in Fig- ure 13.37(a) which may be used as an alternative to the battery symbol.)
Following the procedure
(i) The 3resistance is removed from the circuit as shown in Figure 13.37(b).
(ii) The 123resistance now carries no current.
P.d. across 10resistor= 10
10+5
(24)
=16 V (see Section 13.4(v)).
Hence p.d. across AB, E=16 V
(iii) Removing the source of e.m.f. and replacing it by its internal resistance means that the 20resistance is short-circuited as shown in Figure 13.37(c) since its internal resistance is zero. The 20resistance may thus be removed as shown in Figure 13.37(d) (see Section 13.4 (vi)).
Figure 13.37
From Figure 13.37(d), resistance, r=12
3+10×5 10+5
=12 3+50
15=5
(iv) The equivalent Thévenin’s circuit is shown in Fig- ure 13.37(e), from which
current, I= E
r+R= 16 3+5 = 16
8 =2 A
=current in the 3resistance Problem 11. A Wheatstone Bridge network is shown in Figure 13.38(a). Calculate the current flowing in the 32resistor, and its direction, using Thévenin’s theorem. Assume the source of e.m.f. to have negligi- ble resistance.
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D.c. circuit theory 145
PART
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Figure 13.38
Following the procedure:
(i) The 32 resistor is removed from the circuit as shown in Figure 13.38(b)
(ii) The p.d. between A and C, VAC=
R1
R1+R4
(E)=
2 2+11
(54)=8.31 V The p.d. between B and C,
VBC= R2
R2+R3
(E)= 14
14+3
(54)=44.47 V Hence the p.d. between A and B=44.47−8.31= 36.16 V
Point C is at a potential of+54 V. Between C and A is a voltage drop of 8.31 V. Hence the voltage at point A is 54−8.31=45.69 V. Between C and B is a voltage drop of 44.47 V. Hence the voltage at point B is 54−44.47=9.53 V. Since the voltage at A is greater than at B, current must flow in the direction A to B. (See Section 13.4 (vii)).
(iii) Replacing the source of e.m.f. with a short-circuit (i.e. zero internal resistance) gives the circuit shown in Figure 13.38(c). The circuit is redrawn and simpli- fied as shown in Figure 13.38(d) and (e), from which the resistance between terminals A and B,
r=2×11
2+11+14×3 14+3 = 22
13+42 17
=1.692+2.471=4.163
(iv) The equivalent Thévenin’s circuit is shown in Fig- ure 13.38(f), from which,
current I= E
r+R5 = 36.16
4.163+32=1 A
Hence the current in the 32resistor of Fig- ure 13.38(a) is 1 A, flowing from A to B
Now try the following exercise.
Exercise 42 Further problems on Thévenin’s