Simple Resistive Circuits
Qi Xuan
Zhejiang University of Technology
September 2015
Electric Circuits
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Structure
•  Resistors in Series •  Resistors in Parallel •  The Voltage/Current-­‐Divider Circuit •  Voltage/Current Division •  Measuring Voltage and Current •  Measuring Resistance-­‐The Wheatstone Bridge •  Delta-­‐to-­‐Wye (Pi-­‐to-­‐Tee) Equivalent Circuit Electric Circuits
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A Rear Window Defroster
A Rear Window Defroster
Electric Circuits
A resisEve circuit
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Resistors in Series
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In general, if k resistors are connected in series, the equivalent
single resistor has a resistance equal to the sum of the k
resistances, or
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Resistors in Parallel
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Conductance
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Typical Cases
•  If k = 2, we have •  If R1 =R2 =…… = Rk = R, we have 1/Req = k/R
Req = R/k
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Example #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. Electric Circuits
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R1
i1
v1
i2
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) = 3 A
v1 = i1×R1 = 3×12.8 = 38.4 V
i2 = v1 / (6+10) = 38.4 / 16 = 2.4 A
p = i22×10 = 2.4×2.4×10 = 57.6 W
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Solu5on for Example #1
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The Voltage-­‐Divider Circuits
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The current-­‐Divider Circuit
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Example #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)? Electric Circuits
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Solu5on for Example #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 = vR2 / R= 4802 / 30 = 7680 W
For (c), we have
v = 60×20+vR = 1200+480 = 1680 V
p = 20v = 20×1680 = 33600 W
Electric Circuits
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Voltage/Current Division
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Example #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 cur-­‐ rent 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? Electric Circuits
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Solu5on for Example #3
i
i1
i2
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 = 0.5 A
Rp
i1 = i Rp / 30 = 0.5×10 / 30 A = 166.67 mA
For (c), we have
i2 = i Rp / 60 = 0.5×10 / 60 A = 1/12 A
Electric Circuits
p2 = 50i22 = 50 / 144 W = 347.22 mW
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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 funcEons as a short
circuit in series with the element whose current is being measured. An ideal voltmeter has an infinite equivalent resistance and thus funcEons as
an open circuit in parallel with the element whose voltage is being measured. Electric Circuits
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Digital and Analog Meters
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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. By design, the
deflec5on of the pointer is directly
propor5onal to the current in the
movable coil. The coil is characterized by
both a voltage raEng and a current raEng. For example, one commercially available meter movement is rated at 50 mV and 1
mA. This means that when the coil is carrying 1 mA, the voltage drop across the
coil is 50 mV and the pointer is deflected to its full-scale position.
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DC Ammeter/Voltage Circuit
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Example #4
a)  A 50 mV, 1 mA d'Arsonval movement (le^) is to be used in an ammeter with a full-­‐scale reading of 10 mA. Determine RA. b)  Find the current in the circuit shown (right). c)  If the ammeter (le^) is used to measure the current in the circuit (right), what will it read? Electric Circuits
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Solu5on for Example #4
i
vA
For (a) (left), we have
iA = I - iB=10 mA -1mA = 9 mA
iB
iA
Req
RA = vA / iA = 50 / 9 =5.56 Ω
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= 9.524 mA
Electric Circuits
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Measuring Resistance: The Wheatstone Bridge
Ba`ery
ig = 0
i1 = i3 i2 = ix
i1R1 = i2R2 i3R3 = ixRx
In a commercial Wheatstone bridge, R1 and
R2 consist of decimal values of resistances
that can be switched into the bridge circuit.
R1 / R2 = i2 / i1
R3 / Rx = ix / i3 = i2 / i1
Rx = R2R3 / R1
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Example #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 dissipaEng 250 mW. Can the bridge be le^ in the balanced state without exceeding the power-­‐dissipaEng capacity of the resistors, thereby damaging the bridge? Electric Circuits
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Solu5on for Example #5
i1 i2
p1=i12R1 = 0.022×100 =0.04 W
p2=i22R2 = 0.0022×1000 =0.004 W
p3=i12R3 = 0.022×150 =0.06 W
px=i22Rx = 0.0022×1500 =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
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Delta-­‐to-­‐Wye (Pi-­‐to-­‐Tee) Equivalent Circuits
Δ
Electric Circuits
Y
T
π
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The Δ-­‐to-­‐Y Transforma5on
R1 = (Rab+Rca-Rbc) / 2 = RbRc / (Ra+Rb+Rc)
R2 = (Rab+Rbc-Rca) / 2 = RcRa / (Ra+Rb+Rc)
R3 = (Rbc+Rca-Rab) / 2 = RaRb / (Ra+Rb+Rc)
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How about Y-­‐to-­‐Δ?
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The Y-­‐to-­‐Δ Transforma5on
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Example #6
Use a Y-­‐to-­‐Δ transformaEon to find the voltage v in the circuit shown. Electric Circuits
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Solu5on for Example #6
c
a
Rb
c
b
Ra
a
b
Rc
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
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a
70Ω
28Ω
2A
35Ω
v
105Ω
b
Req = 35||[(28||70)+105||17.5]
= 35||(20+15)
= 17.5 Ω
c
17.5Ω
v = 2Req = 35 V
Summary
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•
•
•
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Series/Parallel resistors Voltage/Current division Digital/Analog voltmeter/ammeter Wheatstone bridge Δ-­‐to-­‐Y/Y-­‐to-­‐Δ transformaEon Electric Circuits
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