Lesson 2 - Ohm's Law and Resisitive Circuits 2.pptx

ELECTRICAL CIRCUITS 1 LECTURE
LESSON 2
Ohm’s Law and Resistive Circuit

 State and apply Ohm’s Law.
 Define power and energy including their units.
 Calculate energy and power.
 Recognizes series, parallel, series-parallel, wye and delta
connections.
 Solve circuits (i.e. find currents and voltages of interest) by combining
resistances in series and parallel.
Learning Objectives
At the end of the lesson, the learner is expected to:

• This law applies to electric to electric conduction
through good conductors and may be stated as
follows :
• “The ratio of potential difference (V) between any two
points on a conductor to the current (I) flowing
between them, is constant, provided the temperature
of the conductor does not change.”
After George Simon
Ohm (1787-1854), a
German
mathematician who
in about 1827
formulated the law
known after his
name as Ohm’s
Law.
Ohm’s Law

where R is the resistance of the conductor between the two points
considered.
• Put in another way, it simply means that provided R is kept constant,
current is directly proportional to the potential difference across the
ends of a conductor. However, this linear relationship between V and I
does not apply to all non-metallic conductors. For example, for silicon
carbide, the relationship is given by V = KI^m where K and m are
constants and m is less than unity. It also does not apply to non-linear
devices such as Zener diodes and voltage-regulator (VR) tubes.
Ohm’s Law

1. A coil of copper wire has resistance of 90 Ω at 20°C and is connected
to a 230-V supply. By how much must the voltage be increased in order
to maintain the current constant if the temperature of the coil rises to
60°C ? Take the temperature coefficient of resistance of copper as
0.00428 from 0°C.
Illustrative Problems: ( Instructor actual discussion)
-

2. Three resistors are connected in series across a 12-V battery. The first resistor has a
value of 1 Ω, second has a voltage drop of 4 V and the third has a power dissipation of 12
W. Calculate the value of the circuit current.
Illustrative Problems: ( Instructor actual discussion)

• Power (P) - it is the rate of doing work. Its units is watt
(W) which represents 1 joule per second.
1 W = 1 J/s
• If a force of F newton moves a body with a velocity of ν
m./s then
Power = F × ν , watt
• If the velocity ν is in km/s, then
Power = F × ν, kilowatt
Power and Energy

Also, in Calculus Based Physics:
, Watts
At any instant:
, Joules
and
, watt-sec or Joules
Note : W is the energy: t is in seconds and P is in watts.
Power and Energy

• Energy (W) - ability of doing electrical work. It is the product of power
obtained and time.
In Electric Circuit,
, watt-sec
Note: P is power, E or V is voltage, I is current, R is resistance and t is time
in sec.
FUNDAMENTALS OF ELECTRICAL CIRCUITS
Power and Energy

Important SI Units:
(i) Kilowatt-hour (kWh) and kilocalorie (kcal)
(ii) Miscellaneous Units
Power and Energy

1. A de-icing equipment fitted to a radio aerial consists of a length of a
resistance wire so arranged that when a current is passed through it,
parts of the aerial become warm. The resistance wire dissipates 1250
W when 50 V is maintained across its ends. It is connected to a d.c.
supply by 100 metres of this copper wire, each conductor of which has
resistance of 0.006 Ω/m.
Calculate:
(a) the current in the resistance wire
(b) the power lost in the copper connecting wire
(c) the supply voltage required to maintain 50 V across the heater itself.
ILLUSTRATIVE PROBLEMS: ( Instructor actual discussion)

2. A factory has a 240-V supply from which the following loads are taken :
Lighting : Three hundred 150-W, four hundred 100 W and five hundred
60-W lamps
Heating : 100 kW
Motors : A total of 44.76 kW (60 b.h.p.) with an average efficiency of 75
percent
Misc. : Various load taking a current of 40 A.
Assuming that the lighting load is on for a period of 4 hours/day, the
heating for 10 hours per day and the remainder for 2 hours/day, calculate
the weekly consumption of the factory in kWh when working on a 5-day
week. What current is taken when the lighting load only is switched on ?

• When some conductors having resistances R1, R2 and R3 etc. are
joined end-on-end as in figure below, they are said to be connected in
series.
• Being a series circuit, it should be remembered that:
(i) current is the same through all the three conductors.
Resistances in Series

(ii) but voltage drop across each is different due to its different
resistance and is given by Ohm’s Law.
(iii) sum of the three voltage drops is equal to the voltage applied across
the three conductors.
where R is the equivalent resistance of the series combination. Also,

• The main characteristics of a series circuit are :
1. same current flows through all parts of the circuit.
2. different resistors have their individual voltage drops.
3. voltage drops are additive.
4. applied voltage equals the sum of different voltage drops.
5. resistances are additive.
6. powers are additive.

Voltage Divider Rule
• Since in a series circuit, same current flows through each of the given
resistors, voltage drop varies directly with its resistance.
Total resistance R = R1 + R2 + R3 = 12 Ω
According to Voltage Divider Rule, various
voltage drops are :
Figure of series network

1. A series circuit consists of three resistors A, B, and C connected to a
120 V source. If the voltage drop across resistor A is 30 volts when
the circuit current is 2A, and Rb = 1.5Rc, Calculate the values of Ra,
Rb, and Rc.
2. The resistors in the voltage-divider circuit shown have a tolerance of 
10%. Find the maximum and minimum value of Vo.

3. A source supplies 120V to the series combination of a 10 ohm
resistance, an 8 ohm resistance and an unknown resistance Rx. The
voltage across the 8 ohm resistance is 20V. Determine the value of the
unknown resistance.
4. Find the voltage Vo for the circuit shown below: Calculate the power at
380V source. Is the power delivering or absorbed?

SOLUTION for NUMBER 4 : By voltage division :
𝑉25 𝑘 = 380 𝑉 (
25 𝑘𝑜ℎ𝑚
25𝑘𝑜ℎ𝑚+75 𝑘𝑜ℎ𝑚
)= 95 V
𝑖 =
95𝑉
25 𝑘𝑜ℎ𝑚
= 0.0038 A
By voltage division to get Vo:
𝐷𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑆𝑜𝑢𝑟𝑐𝑒 = 25,000 𝑥 𝑖 = 25,000 𝑥 0.0038 𝐴 = 95𝑉
𝑉𝑂 = 95 𝑉 (
60𝐾𝑜ℎ𝑚
60 𝑘𝑜ℎ𝑚+40 𝑘𝑜ℎ𝑚
) = 57 volts

• Three resistances, as joined in figure below are said to be connected in
parallel.
(i) potential drop across all resistances is the same
(ii) current in each resistor is different and is given by Ohm’s Law.
(iii) the total current is the sum of the three separate currents.
Figure of parallel network
Resistances in Parallel

• Now,
• The main characteristics of a parallel circuit are :
1. same voltage acts across all parts of the circuit
2. different resistors have their individual current.
3. branch currents are additive.
4. conductance's are additive.
5. powers are additive.

Division of Current in Parallel Circuit
• In the figure below, two resistances are joined in parallel across a
voltage V. The current in each branch, as given in Ohm’s law, is :
Figure of parallel network

• Hence, the division of current in the branches of a parallel circuit is
directly proportional to the conductance of the branches or inversely
proportional to their resistances. We may also express the branch
currents in terms of the total circuit current thus :
• The same approach for 3 or more resistances connected in parallel.

1. Three loads A, B, and C are connected in parallel to a 230 volt source.
Load A takes 9.2 kw. load B takes a current of 60 amp. and load C is a
resistance of 4.6 ohms. Calculate:
(a) the resistances of load A and B,
(b) the total equivalent resistance of the three paralleled loads.
(c) the total current
(d) the total power.

2. A coil wire having a resistance of 3.84 ohms and carrying a current of
0.15 amp. is in parallel with an unknown resistance through which
there is a current of 1.44 amp. Calculate (a) the unknown resistance
(b) the total equivalent resistance.
3. Given the circuit below. Calculate the following:
(a) total equivalent resistance.
(b) total current.
(c) voltage drop across each section.
(d) current through each resistor.
(e) total power taken by the entire circuit.

Figure for problem 3:

4. Use current division to find the current io and use voltage division to
find the voltage vo, for the circuit shown below.
SIMPLIFY THESE RESISTANCES FIRST

SOLUTION :
6.485
1
𝑅𝑇
=
1
10
+
1
80
+
1
24
𝑅𝑇 =
1
0.1542
𝑅𝑇 = 6.485 𝑜ℎ𝑚𝑠

SOLUTION :
6.485
I1
𝐼1 = 8
80
80 + 6.485
= 7.40 𝐴

SOLUTION :
I1= 7.40 A
1
𝑅𝑇
=
1
80
+
1
24
𝑅𝑇 =
1
0.054167
𝑅𝑇 = 18.461
18.861
ohms

SOLUTION :
I1= 7.40 A
Type equation
here.
𝐼2 =7.40(
10
10+18.861
)
18.861
ohms
I2= 2.564 A
𝐼2 = 2.564 𝐴

SOLUTION :
I1= 7.40 A I2= 2.564 A
IO =1.972 A 𝑖𝑜=2.564(
80
80+24
) = 𝟏. 𝟗𝟕𝟐 𝑨
𝑣𝑜 = 2.564 − 1.972 ∗ 30 𝑜ℎ𝑚𝑠
𝑣𝑜 = 17.76 𝑉𝑜𝑙𝑡𝑠

• Circuits combining series and parallel sections with one source of
supply are treated in as much the same way as simple circuits
arrangements.
Figure of series - parallel network
Series-Parallel

1. What is the value of the unknown resistor R in figure below if the voltage
drop across the 500 Ω resistor is 2.5 volts ? All resistances are in ohm.

2. Calculate the effective resistance of the following combination of
resistances and the voltage drop across each resistance when a P.D. of
60 V is applied between points A and B.

3. The equivalent resistance between terminals a and b in the figure
below is Rab = 23 ohms. Determine the value of R2.

SOLUTION for NUMBER 3 :
23 =(
12∗6
12+6
+ 𝑅2) * 80
12 ∗ 6
12 + 6
+ 𝑅2 + 80
+ 7
R2 = 16 ohms

4. Refer to the circuit shown below. With the switch open, we have v2 =
8V. On the other hand, with the switch closed, we have v2 = 6V.
Determine the values of R2 and RL.

• Resistors are sometimes interconnected to form rather complex
networks. Due to this, some common rules applicable to simple series
and parallel circuits cannot be used for the calculation of equivalent
resistances, branch currents, and voltage drops.
∆ - Y and Y - ∆ Transformation

• Considering the connections of resistors below:
• ∆ - Y Transformation:

• For ∆ - Y Transformation : Each of the resistances in the star is equal
to the product of the resistances of the adjacent arms of the delta
divided by the sum of the three delta resistances.
Y-∆ Transformation:
For Y-∆ Transformation : Each of the resistances in delta is equal to the
sum of the products of the resistances in the star, taken two at a time,
divided by the resistance in the opposite leg.

• Drawn as a 4 terminal arrangement of components
T and ( Tee and Pi )

• 2 of the terminals are connects at one node. The node is a
distributed node in the case of the P network.
T and ( Tee and Pi )

1. Referring to the figure in our lecture, the three resistances in a delta-
connected group of resistors are X= 35 ohms, Y = 25 ohms, Z = 40
ohms. Calculate the resistances A, B, and C of the equivalent star.
SOLUTION:
𝐴 =
𝑌𝑍
100
=
25 𝑥 40
100
= 10 𝑜ℎ𝑚𝑠
𝐵 =
𝑋𝑍
100
=
35 𝑋 40
100
= 14 𝑜ℎ𝑚𝑠
𝐶 =
𝑋𝑌
100
=
35 𝑋 25
100
= 8.75 ohms

2. The wiring diagram shown below is known as the Wheatstone Bridge
circuit; when the potential difference between points a and b is zero
the bridge is said to be balanced. For the resistance values given, the
bridge is unbalanced. (a) Calculate the total resistance (b) total current
(c) the current at 110  resistance.

𝐴 =
60 𝑋 80
250
= 19.2 𝑜ℎ𝑚𝑠
SOLUTION:
𝐵 =
60 𝑥 110
250
= 26.4 𝑜ℎ𝑚𝑠
𝐶 =
80 𝑋 110
250
= 35.2 𝑜ℎ𝑚𝑠
Get the equivalent resistance:
𝑅𝑒𝑞 = 19.2 +
35.2+84.8 𝑥 (26.4+33.6)
(120+60)
= 59.2 ohms
(a)

SOLUTION:
(b) 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 𝐼 =
177.6
59.2
= 3 𝑎𝑚𝑝
(c) 𝑇𝑜 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑡ℎ𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 110 − 𝑜ℎ𝑚 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑓𝑖𝑟𝑠𝑡 𝑏𝑒 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒
potential difference between points a and b. To do this, the currents through the branches of the parallel
Circuit of the figure in (a) must be found. These are:
𝐼120 𝑙𝑒𝑓𝑡 𝑏𝑟𝑎𝑛𝑐ℎ =
60
120+60
𝑥 3 = 1 amp
𝐼110 𝑟𝑖𝑔ℎ𝑡 𝑏𝑟𝑎𝑛𝑐ℎ =
120
120 + 60
𝑥 3 = 2 𝑎𝑚𝑝

SOLUTION:
The voltage drop across M is 84.8 x 1 = 84.8 volts, and the voltage drop across N is 33.6 x 2 = 67.2 volts. Therefore,
The potential difference between a and b will be
𝐸𝑎−𝑏 = 84.8 − 67.2 = 17.6 𝑣𝑜𝑙𝑡𝑠
𝐼110 =
17.6
110
= 0.16 𝑎𝑚𝑝
Hence, the current through the 110-ohm resistor is

3. If Ra = 6 , Rb = 9 and Rc = 3 . Convert  to tee network.

Nilsson, James W. 2019. Electric Circuits 11th ed. Amazon [C 621.319 N59
2019(CAL)]
Nahvi, Mahmood 2018. Schaum's Outline of Electric Circuits 7th ed. Amazon [C
621.3192 N14 2018(CAL)]
Kang, James S. 2018. Electric Circuits (MindTap Course List) 1st ed. Amazon [C
621.3192 K13 2018(CAL)]
Bird, John 2014. Electrical Circuit Theory and Technology 6th ed. Amazon [C
621.3192 B53 2014(CAL)]
Alexander, Charles 2017. Fundamentals of Electric Circuits 6th ed. Amazon [C
621.3815 A12 2017(CAL)]
References

Lesson 2 - Ohm's Law and Resisitive Circuits 2.pptx

Recommended

Recommended

More Related Content

Similar to Lesson 2 - Ohm's Law and Resisitive Circuits 2.pptx

Similar to Lesson 2 - Ohm's Law and Resisitive Circuits 2.pptx (20)

Recently uploaded

Recently uploaded (20)

Lesson 2 - Ohm's Law and Resisitive Circuits 2.pptx