EECE 210
Electric
Circuits
Chapter 3 - Simple
Resistive Circuits
Main Topics
• Resistors in Series
• Resistors in Parallel
• The Voltage/Current-‐Divider Circuit
• Voltage/Current Division
• Measuring Voltage and Current
• Measuring Resistance-‐The Wheatstone Bridge
• Delta-Wye Equivalent Circuit
Practical Example of
Resistive Circuits
➢
The Rear Window Defroster
A Rear Window Defroster
A resistive circuit
Introduction
▪ Analytical toolbox:
▪ Ohm’s Law
▪ Kirchhoff’s Law (KVL and KCL)
▪ Complicated Interconnections
▪ Need techniques to simplify the
circuit
▪ Then Apply circuits’ Laws
Resistors in Series
• Same current flows through all the
resistors in series
• If k resistors are connected in series,
the equivalent single resistor has a
resistance equal to the sum of the k
resistances, or
Resistors in Parallel
In a parallel connection of
resistors Apply KCL
Voltage is the same across each resistor
In case of 2 Resistors:
If R1 =R2 =…… = Rk = R, we 1/Req = k/R
Req = R/k
Tip: Req is always smaller than the smallest resistor in the circuit
Resistors Combinations (Examples)
Tip: R3// (short circuit) →
replace the combination by short
circuit (wire): R3 disappears
calculated circuit voltages and currents
Is= 12A, V1=12x6=72V, I1=4A, I2=8A
Resistors Combinations (More Examples)
Notes
▪ Resistors can NOT produce power they only consume it
▪ Current enters from the positive voltage terminal
▪ Short Circuit: Resistance = 0 ==> Voltage difference = 0
▪ Open Circuit: Resistance = ∞ ==> Curent = 0
Solved Problem #1
• For the circuit shown, find
• (a) the voltage v, (b) the power delivered to the circuit
by the current source, and (c) the power dissipated in
the 10 Ω resistor.
Solution for Problem #1
R1 = (10+6)×64 / (10+6+64) = 12.8 Ω
R = (7.2+12.8)×30 / (7.2+12.8+30) = 12 Ω
v = 5R = 60 V
i1 = v / (7.2+R1) = 60 / (7.2+12.8) = 3A
v1 = i1×R1 = 3×12.8 = 38.4 V
i2 = v1 / (6+10) = 38.4 / 16 = 2.4A
p = i 2×10 = 2.4×2.4×10 = 57.6 W
i1
• Let R1 be the equivalent Resistor of the 3 Resistors
in the shaded area circuit above
• Let V1 is the voltage across this shaded area
Voltage Divider Rule
Determine Vk ?
i
+
R1
➢ KCL is satisfied by having a common current i
–
+
through all the resistors
R2
➢ From Ohm’s law: vk = I Rk and vSCR = (R1+
R2+…+Rk) i
vk
Rk
=
Substitute
v SRC R1 + R2 + ... + Rk + ... + Rn
v1
+
–
v2
–
vSRC
+
Rk
vk
–
+
If Vk is in the opposite direction, put a negative
sign in the voltage divider rule
Rn
vn
–
The Voltage- Divider Circuits
Tip: Reduce complexity of circuit
Then Apply voltage divider rule
Voltage Divider Rule Example
▪ Determine V2 and V1
▪
Reduce complexity of circuit
▪ Parallel Resistors (15//3) in series with
the parallel combination of (9//45)
▪
Apply voltage divider rule
Current Divider Rule
▪ Determine ij
v
ij =
Rj
ij = i
Req (//)
Rj
▪ If ij is in the opposite direction, put a negative sign in the current divider rule
The Current-Divider Circuit
Determine i1
- KVL is satisfied in every loop v= v1=v2
- From Ohm’s law v = i1R1 and v = i2R2
- Find Req of the 2 // resistors R1 and R2
- Apply KCL to check iS = i1 + i2
Solved Problem #2
a) Find the value of R that will cause 4 A of current to
flow through the 80 Ω resistor in the circuit shown.
b) How much power will the resistor R from part (a)
need to dissipate?
c) How much power will the current source generate
for the value of R from part (a)?
Solution for Problem #2
i
vR
For (a), we have
i = 20R / (40+80+R) = 4 A
R = 30 Ω
For (b), we have
vR = (40+80)i = 120×4 = 480 V
pR = v 2 / R= 4802 / 30 = 7680 W
R
For (c), we have
v = 60×20+vR = 1200+480 = 1680 V
p = 20v = 20×1680 = 33600 W
Solved Problem #3
a) Use voltage division to determine the voltage v0 across the
40 Ω resistor in the circuit shown.
b) Use v0 from part (a) to determine the current through the 40
Ω resistor and use this current and current division to
calculate the current in the 30 Ω resistor.
c) How much power is absorbed by the 50 Ω resistor?
Solution for Problem #3
i
For (a), we have
Req = 40+(20||30||(50+10))+70
= 40+10+70=120 Ω
vo = 60×40 / Req = 20 V
For (b), we have
i = vo / 40 = 20 / 40 = 0.5A
Rp
i1 = i Rp / 30 = 0.5×10 / 30 A = 166.67mA
For (c), we have
i2 = i Rp / 60 = 0.5×10 / 60 A = 1/12A
p2 = 50i 22 = 50 / 144 W = 347.22 mW
Measuring Voltage and Current
▪ An ammeter is an
instrument designed to
measure current
▪ A voltmeter is an
instrument designed to
measure voltage
▪ An ideal ammeter has an equivalent resistance of 0 Ω and functions as a
short circuit in series with the element whose current is being measured.
▪ An ideal voltmeter has an infinite equivalent resistance and thus functions
as an open circuit in parallel with the element whose voltage is being
measured.
Resistivity Practical Perspective
▪ Resistors: property of material regarding current flow
▪ Resistivity (ρ): Different materials allow the current
to flow differently
▪
Insulators: very high resistivity
▪
The lower the resistivity, the better is the wire
l
R=
A
Measuring DC Voltage and Current
▪ DC: Direct Current
▪
All power sources we have studied so far are DC dependent
and independent sources
▪ Measuring Devices
▪
Ammeter: An ammeter is an instrument that measures current
through a circuit element when inserted in series with that
element
▪
Voltmeter: A voltmeter is an instrument that measures
the voltage across a circuit element and inserted in
parallel with that element
Measuring DC Voltage and Current
▪ Ammeter
Ideally zero resistance (very small)
▪ Voltmeter
▪
Ideally infinite resistance (very big)
▪ Two Types of Meters
▪
Digital Meters
▪
Easy to connect
▪
Easy and precise readout
▪
Analog meters
▪
Screen with pointer
▪
d’Arsonval meter movement
▪
d’Arsonval Meter Movement
• d'Arsonval meter movement consists
of a movable coil placed in the field of
a permanent magnet.
• When current flows in the coil, it
creates a torque on the coil, causing it
to rotate and move a pointer across a
calibrated scale.
• The deflection of the pointer is
directly proportional to the current in
the movable coil.
• The coil is characterized by both a
voltage rating and a current rating.
DC Ammeter/Voltage Circuit
➢ Commercially available meter
movement is rated at 50 mV and 1
mA.
➢ When the coil is carrying 1 mA, the
voltage drop across the coil is 50 mV
and the pointer is deflected to its fullscale position.
Typical Digital and Analog Meters
Solved Problem #4
a) A 50 mV, 1 mA d'Arsonval movement is to be used in an ammeter
with a full scale reading of 10 mA. Determine RA.
b) Find the current in the below shown circuit shown.
c) If the ammeter is used to measure the current in the below shown
circuit, what will it read?
Solution for Problem #4
i
vA
For (a) (left), we have
iA = I - iB=10 mA -1mA = 9 mA
iB
iA
RA = vA / iA = 50 / 9 =5.56 Ω
Req
For (b) (right), we have
i = 1 / 100 A = 10 mA
For (c), we have
Req = vA / i = 50/ 10 = 5 Ω
Req
i = 1 / (100+Req) = 1 / 105 A= 9.524mA
Measuring Resistance - The Wheatstone Bridge
Operation Principle
a
▪
▪
▪
▪
▪
Many different circuit configurations are used to measure
resistance. “The Wheatstone Bridge” is one of these
configurations
“The Wheatstone Bridge” is used to precisely measure
resistances of medium values, that is, in the range of 1 Ohm
to 1 Mega Ohm with accuracies on the order of ± 0.1 %.
Determine the value of an unknown resistance Rx in terms of
known resistances.
R3 is varied until the ammeter current becomes zero, which
is the condition for bridge balance
Balanced Bridge
▪ Nodes b and c are at the same voltage and no current
flows between them.
▪ The branch “bc” could be open circuited or short
circuited
R1
VSRC
+
–
R2
A
b
R3
I=0
Rx
d
c
Measuring Resistance: The Wheatstone Bridge
C
ig is zero (no current through circuit branch
Battery
ig = 0 “ab”) means that the bridge is balanced AND
i1 = i3 i2 = ix
i1R1 = i2R2 i3R3 = ixRx (At bridge balance
Vcb=Vca & Vad=Vbd)
d
Tip: R1 and R2 are decimal resistances and
can be switched into the bridge circuit.
R1 / R2 = i2 / i1
R3 / Rx = ix / i3 = i2 / i1
Rx = R2R3 /R1
Solved Problem #5
The bridge circuit shown is balanced when R1=100 Ω, R2 =1000
Ω, and R3 =150 Ω. The bridge is energized from a 5 V dc source.
a) What is the value of Rx?
b) Suppose each bridge resistor is capable of dissipating 250
mW. Can the bridge be in the balanced state without
exceeding the power- dissipating capacity of the resistors,
thereby damaging the bridge?
Solution for Problem #5
i1 i2
2
2
p1=i1 R1 = 0.02 ×100 =0.04 W
p2=i22R2 = 0.0022×1000 =0.004 W
2
2
p3=i1 R3 = 0.02 ×150 =0.06 W
2
2
px=i2 Rx = 0.002 ×1500 =0.006 W
For (a), we have
Rx = R2R3 / R1 = 1000×150 / 100 = 1500
For (b), we have
i1= v / (R1+R3)= 5 / 250 = 0.02 A
i2= v / (R2+Rx)= 5 / 2500 = 0.002 A
Delta-to-Y Transformation
▪ Find Req
Series?
▪ Need to simplify the circuit BUT simple
series and Parallel does not work
▪ USE Delta-to-Y Transformation
▪ Also referred to as Pi-to-Tee equivalent
Parallel?
Delta-to-Wye (Δ to Y) Equivalent Circuits
Δ
T
π
27
Delta-to-Wye Transformation
▪
Delta (pie) Connection
▪
Wye (Tee) Connection
Delta-to-Wye Transformation
▪ Y to ∆
▪ ∆ to Y
Rc
a
b
R1
R2
R3
Rb
Ra
c
Delta-to-Wye Transformation (Example I)
▪ Find the current and power supplied by the 40 V in the circuit
Solved Problem #6
Use Y to D Transformation to find voltage v in the circuit
c
a
Rb
c
a
Ra
2A
b
a
70Ω
28Ω
35Ω
v
c
b
Rc
105Ω
17.5Ω
Ra = (20×10+20×5+10×5) / 20 =17.5 Ω
Rb = (20×10+20×5+10×5) / 5 = 70 Ω
Rc = (20×10+20×5+10×5) / 10 = 35 Ω
b
Req = 35||[(28||70)+105||17.5] = 35||(20+15) =17.5 Ω
v = 2Req = 35 V
Review Problems
▪ Determine i2?
Review Problems
▪ Determine P dissipated in the 6Ω resistor ?
I0=8A, i6=3.2A, P=61.44W
Review Problems
▪ Find vo when
▪
Load is 150 kΩ, 133.33V
▪
No load is connected 150V
▪
Load is short circuited accidently, Find
power dissipated by 25 kΩ, 1.6W
▪
Find maximum power dissipated in 75
kΩ resistor 0.3W
Review Problems
▪ Find V
▪ Hint: Use Y-to-delta transformation:
▪ V= 35V
Chapter Review- What We have Learned
• Series/Parallel resistors
• Voltage/Current division
• Digital/Analog voltmeter/ammeter
• Wheatstone bridge
• Delta to Wye to Delta Transformation