1. Kirchhoff's laws (KCL and KVL) form the foundation for electric circuit analysis. KCL states that the algebraic sum of currents entering a node is zero, while KVL states that the algebraic sum of voltages around a closed loop is zero. 2. A node is the point of connection between two or more branches, while a loop is any closed path in a circuit without repeating nodes. Kirchhoff's laws can be applied to nodes, closed surfaces, and loops. 3. Ohm's law defines the relationship between current and voltage in a resistor as V=IR. Resistors can be connected in series or parallel, and equivalent resistances can be calculated using their individual

1. Chapter-2: Basic Laws
2.1: Introduction
• Knowledge of basic concepts, acquired in last chapter, itself is not
enough to meet the objective of executing circuit analysis,.
• To investigate circuit variables like voltage, current or power requires
deep understanding of fundamental laws that govern electric circuits.
• Basic laws like the Ohm’s and Kirchhoff’s laws, form the foundation
upon which electric circuit analysis is built.
• Additionally the familiarity with techniques commonly applied in
circuit design and analysis is also essential to perform circuit analysis.

2. • Such techniques include combining resistors in series or parallel,
voltage division, current division, and delta  wye transformations.
• To begin with, these laws and techniques are applied to resistive
circuits only and will be later expanded to other circuit elements.
2.2: Ohm’s Law
• Materials in general have a characteristic behavior of resisting the
flow of electric charge.
• This property or ability of materials to resist current, is known as
resistance and is represented by the symbol R.

3. • Resistance of any material depends directly on its length (l) and
indirectly on its cross-sectional area (A), as shown in Fig. 2.1(a).
• We can represent resistance of any material in mathematical form as,
• (2.1).
• Variable  (Rho) is known as resistivity
of material, measured in ohm-meters.
• Good conductors, e.g. gold & copper
have low  values, while insulators, e.g.
mica & paper, have high  values.

4. • Table 2.1 provide values of  for some common materials used as
conductors, insulators and semiconductors.
• Circuit element used to model the current-resisting behavior of a
material is the resistor.
• Class Work: Distinguish between resistance, resistivity & resistor?

5. • Resistors used in electrical circuits are usually made
from metallic alloys and carbon compounds.
• Circuit symbol for the resistor is shown in Fig. 2.1(b),
where R stands for the resistance of the resistor.
• The resistor is the passive element i.e. it dissipates energy as heat.
• The relationship between current and voltage for a resistor is best
defined by what is known as Ohm’s law.
• Ohm’s law states that the voltage v across a resistor is directly
proportional to the current i flowing through it i.e. v  i (2.2).

6. • Selecting resistance R of a resistor as constant of proportionality, the
mathematical form of Ohm’s law becomes
• v = iR (2.3).
• Resistance R of any circuit element denotes its ability to resist the
flow of electric current; and it is measured in ohms ( ).
• We may deduce from Eq. (2.3) that R = (2.4).
• Thus 1  = 1 V/A.
• To apply Ohm’s law as stated in Eq. (2.3), pay careful attention to the
current direction and voltage polarity.

7. • Direction of current i & polarity of voltage v must conform
with passive sign convention, as shown in Fig. 2.1(b).
• Conformation implies that current flows from a higher
potential to a lower potential and Ohm’s Law v = iR applies.
• Non-conformation to passive sign convention means the current flows
from a lower to a higher potential, and Ohm’s law v = – iR applies.
• Since the value of R can range from zero to infinity, it is important
that we consider the two extreme possible values of R.
• Element with R = 0 is called a short circuit, as shown in Fig. 2.2(a).

8. • For a short circuit, v = iR = 0 (2.5).
• Eq. (2.5) shows that the voltage is zero
but the current could be anything.
• In practice, a short circuit is usually a
connecting wire assumed to be a perfect
conductor.
• Thus, a short circuit is a circuit element
with resistance approaching zero.
• Similarly, an element with R =  is an
open circuit, as shown in Fig. 2.2(b).

9. • For an open circuit, (2.6).
• Eq. (2.6) shows that current is zero though voltage could be anything.
• Thus, open circuit is circuit element with resistance reaching infinity.
• A resistor is available either as fixed or variable element.
• Most resistors are of fixed type i.e. their resistance remains constant.
• The two common types of fixed resistors (wire wound and
composition) are shown in Fig. 2.3.
• The composition resistors are used when large resistance is needed.
• The circuit symbol that we saw in Fig. 2.1(b) is for a fixed resistor.

10. • Variable resistors have adjustable resistance.
• The symbol for a variable resistor is shown in
Fig. 2.4(a).
• A common variable resistor is known as a potentiometer or pot for
short, with the symbol shown in Fig. 2.4(b).
• The pot is a three-terminal element with a sliding contact or wiper.
• Sliding the wiper, the resistances between it and fixed terminals vary.
• Like fixed resistors, variable resistors can be of either wire wound or
composition type, as shown in Fig. 2.5.

11. • Although resistors like those in Figs. 2.3 and 2.5 are used in circuit
designs, today most circuit components including resistors are either
surface mounted or integrated, as typically shown in Fig. 2.6.
• Not all resistors obey Ohm’s law, resistor that
obeys Ohm’s law is known as a linear resistor.
• It has a constant resistance and its i-v
characteristic is illustrated in Fig. 2.7(a):
• A nonlinear resistor does not obey Ohm’s law.
• Its resistance varies with current and its i-v
characteristic is typically shown in Fig. 2.7(b).

12. • Examples of devices with nonlinear resistance are light bulb & diode.
• Please note that all practical resistors often exhibit nonlinear behavior
under certain conditions.
• Nevertheless, we will assume in this course that all elements actually
designated as resistors are linear.
• Another useful quantity in circuit analysis is the reciprocal of
resistance R, known as conductance and denoted by G:
• G = (2.7).
• Conductance is measure of how well an element will conduct current.

13. • SI unit of conductance is Siemens (S) and 1 S = 1 A/V (2.8).
• Same resistance can be expressed in Ohms or Siemens.
• For example, 10  is the same as 0.1 S.
• From Eq. (2.7), i = Gv (2.9).
• Power dissipated by a resistor can be expressed in terms of R & G i.e.
• p = vi = i2R = v2/R (2.10).
• p = vi = v2G = i2/G (2.11).
• We should note two things from Eqs. (2.10) and (2.11):
1. Power in resistor is nonlinear function of either current or voltage.

14. 2. Since R & G are positive, power in resistor is always positive.
• Thus, a resistor always absorbs power from the circuit.
• This confirms the idea that a resistor is a passive element, incapable
of generating energy.
• Example 2.1: An electric iron draws 2 A at 120 V. Find its resistance?
• Applying ohm’s law R = v/i = 120/2 = 60 .
• Home Work: P_Problem 2.1.
• Example 2.2: In the circuit shown in Fig. 2.8, calculate the current i,
the conductance G, and the power p.

15. • The voltage across the resistor is the same as
the source voltage (30 V). How?
• Because the resistor and the voltage source are both connected to the
same pair of terminals.
• Hence, the current i = v/R=30/5000 = 6 mA.
• Thus conductance G = 1/R = 1/5000 = 0.2 mS.
• Class Work: Compute power using p = i2R and p = v2/R?
• In either case answer remains p = iv = 0.006 x 30 = 180 mW
• Home Work: P_Problem 2.2

16. • Example 2.3: Consider a voltage source of 20 sin t V connected
across a 5 k resistor. Find the current through the resistor and the
power dissipated.
• , therefore;
• p = vi = (20 sin t)(4 sin t) = 80 sin2 t mW. Check by p = i2R = v2/R
• Home Work: P_Problem 2.3.
2.3: Nodes, Branches and Loops
• Elements of an electric circuit can be interconnected in several ways.
• To do so we must understand basic concepts of network topology.

17. • What is the difference between a network and a circuit?
• Principally the term network & circuit mean the same thing.
• Though Network is interconnection of elements & devices but circuit
is a network providing one or more closed paths for current flow.
• The convention, when addressing network topology, is to use the
word network rather than circuit.
• In network topology, we study properties relating to the placement of
elements in the network and geometric configuration of the network.
• A network topology is stated in terms of branches, nodes and loops.

18. • Branch represents any two-terminal active or passive circuit element.
• The circuit in Fig. 2.10 has five branches, namely, the 10 V voltage
source, the 2 A current source, and the three resistors.
• A node is the point of connection between two or more branches.
• A node is usually indicated by a dot in a circuit.
• If a short circuit (a connecting wire) connects two nodes, the two
nodes constitute a single node.
• The circuit in Fig. 2.10 has three nodes
a, b and c.

19. • Notice that the three points that form node b are connected by
perfectly conducting wires and therefore constitute a single point.
• The same is true of the four points forming node c.
• To demonstrate that the circuit in Fig. 2.10 has only three nodes, the
circuit is redrawn in Fig. 2.11.
• The two circuits in Figs. 2.10 and
2.11 are identical, however, for the
sake of clarity, nodes b and c are
spread out with perfect conductors.

20. • A loop is any closed path in a circuit where no node appears twice.
• A loop is formed by starting at a node, passing through a set of nodes,
and returning to the starting node without passing through any node
more than once. (except the start and finish nodes that are same)
• A loop is said to be independent if it contains at least one branch,
which is not a part of any other independent loop.
• Independent loops produce independent sets of equations.
• It is possible to form an independent set of loops where one of the
loops does not contain any such a branch i.e. non-shared branch.

21. • In Fig. 2.11, loop abca with the 2  resistor is first independent loop.
• Second independent loop comprises of 3  resistor & current source.
• To complete the set of three independent loops, 2  parallel with 3 
resistor, which has no unshared branch, could be taken as third loop.
• A network with b branches, n nodes, and l independent loops will
satisfy the fundamental theorem of network topology i.e.
• OR 5 = 2 + 4 – 1 (2.12).
• Not in book, but formula b – n + 1 can also be
used to determine needed simultaneous equations
to solve a circuit by loop or mesh analysis.

22. • What is a mesh? Mesh is a loop where NO closed path exists inside it.
• Circuit in Fig 2.11 has three meshes. Can you identify them?
• Remember; every mesh is a loop, but every loop may not be mesh.
• We need not remember formulae to acquire required set of equations.
• While drawing the mesh or loop, just ensure that each circuit element
is counted at least once in one of the selected loop or mesh.
• Two circuit elements are said to be in series if they exclusively share
a single node and consequently carry the same current.

23. • Two or more elements are in parallel if they are connected to the
same two nodes and consequently have the same voltage across them.
• Elements may be connected in a way that they are neither in series
nor in parallel e.g. 5 & 2  resistors in circuit shown in Fig. 2.10
• Class Work: Identify the elements connected in series & parallel?
• Note that the voltage source and the 5  resistor are in series while 2
& 3  resistors and source are in parallel.
• Self Study: Example 2.4 and P_Problem 2.4.

24. 2.4: Kirchhoff’s Laws
• Ohm’s law by itself is not sufficient to analyze every circuit.
• However, coupled with Kirchhoff’s laws, it does provide a amply
powerful set of tools for analyzing a large variety of electric circuits.
• Kirchhoff’s laws are formally known as Kirchhoff’s Current Law
(KCL) and Kirchhoff’s Voltage Law (KVL).
• Kirchhoff’s first law (KCL) is based on the law of conservation of
charge, which requires that the algebraic sum of charges within a
system cannot change.

25. • Kirchhoff’s Current Law (KCL) states that the algebraic sum of all
currents entering a node (or a closed boundary/surface) is zero.
• Mathematically, KCL is expressed (2.13).
• Where N is the number of branches connected to the node and in is the
nth current entering (or leaving) the node.
• By this KCL definition, currents entering a node are deemed positive,
while currents leaving may be taken as negative or vice versa.
• Self Study; Students should go through the proof of KCL, given on
page 38 of the book, in their own time

26. • Applying KCL on node of Fig. 2.16 give,
•
• By rearranging the terms, yield,
•
• This provide yet another way of describing the KCL and what is that?
• The sum of the currents entering a node is equal to the sum of the
currents leaving the node. What are the other two KCL definitions?
• KCL also applies to a closed boundary (or surface). What is closed surface?
• In two dimensions, a closed boundary is the same as a closed path.

27. • A closed surface can also be defined as a set of circuit elements under
a closed path drawn on circuit that are interconnected electrically.
• Apply KCL to closed surface, illustrated in the circuit of Fig. 2.17?
• Answer: Currents entering/leaving a closed surface equal currents
leaving/entering it OR sum of currents entering/leaving it is zero.
• Application of KCL allow to combine
multiple current sources in parallel circuit.
• By KCL; total current in parallel circuit
must equal sum of currents by all sources.

28. • For example; consider parallel current
sources of circuit in Fig. 2.18(a).
• Equivalent (combined) current source can
be found by applying KCL to node a i.e.
• IT = I1 – I2 + I3 (2.18).
• Equivalent circuit is as in Fig. 2.18(b).
• Sources are equivalent if they have same terminal iv-characteristics.
• Kirchhoff’s Voltage Law (KVL) is the second Kirchhoff’s law that is
based on the principle of conservation of energy.

29. • Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of all
voltages around a closed path (or loop) is zero.
• Expressed mathematically, KVL states that
• (2,19).
• Where M is the number of voltages in the loop (or the number of
branches in the loop) and is the mth voltage.
• Consider the circuit shown in Fig. (2.19).
• To list loop voltages, start at any branch
and traversed the loop in either clock wise
or anticlockwise direction.

30. • KVL applies to each loop independently.
• While writing the KVL equation, every voltage in loop is assigned
polarity of the element’s terminal polarity encountered first in a loop.
• Going clockwise around the loop; the resulting KVL equation would
be – v1 + v2 + v3 – v4 + v5 = 0 (2.20).
• Rearranged: v2 + v3 + vs = v1 + v4 (2.21).
• Sum of voltage drops = Sum of voltage rises (2.22).
• This represents an alternative form of defining the KVL.
• Note; going counterclockwise will yield same result as in Eq. (2.21).

31. • Except for the reversed signs, result also remain same as in Eq. (2.20).
• When voltage sources are connected in series, KVL can be applied to
obtain the equivalent voltage source.
• The combined or equivalent voltage is the algebraic sum of the
voltages of the individual sources as shown in Fig. (2.20).
• A circuit cannot contain two different voltages V1 and V2 in parallel or
KVL is violated unless V1 = V2.
• Can two different currents flow
simultaneously in a series circuit?

32. • That is not possible, unless I1 = I2; otherwise KCL will be violated.
• When ever current flows through resistor, a voltage develops across it,
which is referred to as voltage drop or load voltage.
• Voltage drop is part of the source voltage but both are not same.
• Source voltage provides the electrical energy to move electrons while
voltage drop converts the electrical energy into another energy form.
• Voltage drop across a load has a polarity, just like the voltage source.
• But polarity of voltage drop indicates only direction of current flow.
• Load voltages always equal source voltages in electrical circuit.

33. • Example 2.5: For the circuit in
Fig. 2.21(a), find voltages v1
and v2?
• Applying the Ohm’s law: v1 = 2i and v2 = – 3i (2.5.1).
• Applying KVL around the loop in Fig. 2.21(b) yield KVL equation.
• – 20 + v1 – v2 = 0 (2.5.2).
• Substituting values from Eq. (2.5.1) yield – 20 + 2i – (– 3i) = 0.
• Simplifying reveal i = 4 A.
• Replacing i = 4 A in Eq. (2.5.1) give back v1 = 8 V and v2 = 12 V

34. • Home Work: P_Problem 2.5.
• Example 2.6: Determine vo and i in the circuit shown in Fig. 2.23(a)?
• Applying KVL around the loop as shown in Fig. 2.23(b)..
• – 12 + 4i + 2vo – 4 – (– 6i) = 0 (2.6.1).
• Since vo = – 6i (Ohm’s law), replacing its value in Eq. 2.6.1 reveal,
• – 16 + 4i – 12i + 6i = 0, solving it reveal i = – 8 A and vo = 48 V.

35. • Home Work: P_Problem 2.6.
• Example 2.7: Find io and vo in the parallel circuit shown in Fig. 2.25?
• Applying KCL to node a, yield;
• For the 4 resistor, Ohm’s law gives
• Home Work: P_Problem 2.7.
• Example 2.8: Find currents & voltages in circuit of Fig. 2.27(a)?

36. • By Ohm’s law, v1 = 8i1 , v2 = 3i2 and v3 = 6i3 (2.8.1).
• Application of KCL at node a, yield; i1 – i2 – i3 = 0 (2.8.2).
• KVL applied to loop -1 reveal;
• But v1 = 8i1 and v2 = 3i2, therefore – 30 + 8i1 + 3i2
• Hence (2.8.3).

37. • Applying KVL to loop 2; v2 = v3 (2.8.4).
• Replacing value of v2 and v3 from Eq. (2.8.1) leads to,
• 6i3 = 3i2 that leads to i3 = (2.8.5).
• Substituting i1 & i2 (Eqs. 2.8.3 & 2.8.5) into KCL Eq. (2.8.2) reveal,
• and once solved i2 = 2 A.
• Using the value of i2, in Eqs. (2.8.1) to (2.8.5) reveal;
• i1 = 3 A, i3 = 1 A, v1 = 24 V, v2 = 6 V and v3 = 6 V.
• Home Work: P_Problem 2.8.

38. 2.5: Series Resistors and Voltage Division
• Single-loop circuit of Fig. 2.29, contains only two resistors in series.
• Equivalent resistance of any number of resistors connected in series is
sum of individual resistances i.e. Req = = R1 + R2 +…+Rn.
• Thus circuit of Fig 2.29 can be redrawn as shown in Fig 2.30

39. • Since the same current i flows in both of them, applying Ohm’s law to
each of the resistors, we obtain v1 = iR1 & v2 = iR2 (2.24).
• Notice that the source voltage v is divided among the resistors in
direct proportion to their resistances; i.e. larger the resistance, the
larger the voltage drop.
• Thus voltage drop across each resistor in single loop can also be
computed without knowing the series current by voltage division i.e.
• (2.30).
• This is called the principle of voltage division, and the circuit in Fig.
2.29 is called a voltage divider.

40. • In general, if a voltage divider has N resistors in series with the source
voltage v, then voltage drop across the nth resistor ( Rn) will be;
• (2.32).
2.6: Parallel Resistors and Current Division
• Consider the circuit in Fig. 2.31,
where two resistors are connected
in parallel and therefore have the
same voltage “v” across them.
• By Ohm’s law,

41. • Thus; and applying KVL
• We can also find the circuit’s equivalent resistance and apply Ohm’s
law to directly find the total current in the circuit i.e. i = v  Req.
• Equivalent resistance (Req) of the resistors in parallel is given by
reciprocal formula, described in three ways i.e.
• (2.37).
• (2.38).
• (2.39).
• Proof of same is given on page-45 for those who are interested.

42. • Equivalent conductance of parallel resistors is obtained the same way
as the equivalent resistance of series resistors and vice versa i.e.
• In parallel circuit; Geq = G1 + G2 + G3 + … + GN (2.40).
• Thus circuit in Fig. 2.31 can be replaced with the
simple circuit shown in Fig. 2.32.
• Bur if the circuit of Fig. 2.31 has current instead of a voltage source
then currents i1 & i2 cant be determined using Ohm’s law.
• Current i is shared by parallel resistors in inverse proportion to their
resistances, thus, in two branch circuit, such as the one in Fig. 2.31;

43. • (2.43).
• This is known as the principle of current division (or rule), and the
circuit in Fig. 2.31 is known as a current divider.
• For a current divider circuit with N-resistors in parallel with the
source current i, then current in nth Resistor ( Rn) is given by,
• in = Req  Rn, where Req is given by Eq. (2.38).
• Similarly, if a current divider has N conductors in parallel with the
source current i, then current in nth conductor ( Gn) is given by,
• (2.45).

44. • A practical circuit is usually a combination of series as well as parallel
connections as shown in Fig. 2.34. How to find Req in such circuits?
• Such circuits can always be solved, reducing two resistors at a time,
into their equivalent resistor as per their series/parallel connectivity.
• Example 2.9: Find Req in circuit shown in Fig 2.34?
• Req = (5 + 1) || {2 + (6 || 3)} + 4 + 8 = 14.4 .
• Home Work: P-Problem 2.9.
• Example 2.10: Find Rab in circuit in Fig. 2.37?
• Rab = (((5 + 1) || (4 || 12)) +1) || ((6 || 3) + 10) = 11.2 .
• Home Work: P_Problem 2.10.

45. • Example 2.11; Find Geq for the circuit
in Fig. 2.40(a)?
• Geq = ((12 + 8) || 5) + 6 = 10 S.
• We can also do it the hard way.
• First convert all conductances into
resistances as shown in Fig, 2.40(c).
• Next compute
• Solving it, Req =
• Hence,
• Home Work: P_Problem 2.11.

46. • Example 2.12: Find iO and vO in the circuit shown in Fig. 2.42(a).
Also calculate the power dissipated in the 3 resistor.
• Adding 6 || 3 = 2  reduces the circuit to simple circuit of Fig
2.42(b) allowing application of Ohm’s law; i = 12 ÷ (4 + 2) = 2 A.
• Hence, vo = 2i = 2 x 2 = 4 V.
• Applying voltage division also yield same
answer. How?
•
• Applying Ohm’s law; io = 4 ÷ 3 = A.

47. • Applying current division verifies this answer. How?
•
• Power dissipated in 3  resistor =
• What are the other options to find this power?
• Home Work: P_Problem 2.12.
• Example 2.13: For the circuit shown in Fig.
2.44(a), determine: (a) the voltage vO (b) the
power supplied by the current source, (c) the
power absorbed by each resistor.

48. • Combining resistors 6 + 12 = 18  reduces the circuit to Fig 2.44 (a).
• Applying the current division rule;
•
•
• Notice that voltage across the 9 k & 18 k resistors is same. Why?
• Therefore;
• Power supplied by source = voio = (180)(30 x 10–3) = 5.4 W.
• p12k =
• p6k

49. • p9k
• Also p9k = voio = (180)(20 x 10–3) = 3.6 W.
• Notice that the power supplied (5.4 W) equals the power absorbed
(1.2 + 0.6 + 3.6 = 5.4 W). This is one way of checking results.
• Home Work: P_Problem 2.13.
2.7: Wye Delta Transformations
• Many practical circuit, such as
the bridge circuit of Fig. 2.46,
have resistors that are neither in
parallel nor in series.

50. • Such circuits can be simplified by using three-terminal equivalent
networks, called the wye (Y) or tee (T) network shown in Fig. 2.47 and
the delta () or pi () network shown in Fig. 2.48.
• These networks occur by themselves or as part
of a larger network & are used in 3 networks,
electrical filters, and matching networks.

51. • For wye  delta transformation,
superimpose a target network on the
existing donor network as shown in Fig
2.49 and find its equivalent resistances.
• In   Y conversion, each equivalent resistance in target Y network is
obtained from donor  network by dividing product of the resistors in
the two adjacent  branches by the sum of the three  resistors.
• For example; applying this rule to Y target on donor  in Fig. 2.49,
• , &

52. • In Y   conversion, equivalent resistance in the target  network is
obtained from donor Y network by dividing the sum of all possible
products of Y resistors taken two at a time, by the opposite Y resistor.
• For example; from Fig. 2.49,
• Similarly;
• Class Work: Find 3rd  resistor?
• The Y and  networks are said to be balanced when,
• (2.56).
• Under these conditions, conversion formulas become,

53. • (2.57).
• Example 2.14: Convert the  network in Fig. 2.50(a) to Y network?
• Superimpose a Y over  and compute equivalent Y resistances i.e.
• R1 = (RbRc)  (Ra + Rb + Rc) = (15 + 10 + 25)  50 = (250  50) = 5 .
• R2 = (RcRa)  (Ra + Rb + Rc) = (375  50) = 5 .
• R3 = (RaRb)  (Ra + Rb + Rc) = (150  50) = 3 .
• Resulting equivalent Y network
is shown in Fig. 2.50(b).
• Home Work: P_Problem 2.14.

54. • Class Work: Alter shown  into Y network?
• Shown  network was solved in Example
2.14 and has only changed variables.
• Place target Y on donor  network, assign
variables randomly and compute Req,
• Rbn = (R1R2)  (R1 + R2 + R3) = 250  50 = 5 .
• Rcn = (R1R3)  (R1 + R2 + R3) = 375  50 = 5 .
• Ran = (R2R3)  (R1 + R2 + R3) = 150  50 = 3 .
• Note that resulting equivalent Y-network remains unchanged.

55. • Example 2.15: Obtain equivalent resistance Rab
in circuit of Fig. 2.52 and use it to find current i.
• Circuit contains multiple Y and  networks.
• Transforming just one of these will simplify the circuit.
• Taking donor Y network comprising R1 = 10  , R2 = 20  , R3 = ,
• = 35 .
• .
• .

56. • Replacing Y with equivalent  network the circuit is transformed into
series-parallel circuit as shown in Fig. 2.53(a).
• Combining the three pairs of resistors in parallel, we obtain
• 70 || 30 = 21, 12.5 || 17.5 = 7.292  and15|| 35 = 10.5 .
• Hence,
• Resulting equivalent circuit is shown in Fig. 2.53(b).

57. • Therefore,
• To check the answer, transform a donor  into target Y network
• Select “can” delta in Fig 2.52 comprising of Rc = 10  , Ra = 5  and
Rn = 12.5  resistors as donor and transform it into target Y network.
•
•
•
• Substituting target Y network in circuit
changes it into one shown in Fig 2.53(c).

58. • We can now solve it as normal series-parallel circuit.
• Class Work: Find Rab in circuit of Fig 2.53(c)? (Rab = 9.631 )
• This now leads to
• Note that using two variations of transformation leads to the same
results, which represents a very good check on the answer.
• Home Work: P_Problem 2.15.
• Self Study: Sec 2.8 Applications, Sec 2.9 Summary and solve all
chapter end review questions.
• Home Work: All chapter end problems that are similar to the problems
we solved in this chapter.