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Voltage and Current Division. Objective of Lecture. Explain mathematically how a voltage that is applied to resistors in series is distributed among the resistors. Chapter 2.5 in Fundamentals of Electric Circuits Chapter 5.7 Electric Circuit Fundamentals - PowerPoint PPT Presentation
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Objective of LectureExplain mathematically how a voltage that is applied
to resistors in series is distributed among the resistors.Chapter 2.5 in Fundamentals of Electric CircuitsChapter 5.7 Electric Circuit Fundamentals
Explain mathematically how a current that enters the a node shared by resistors in parallel is distributed among the resistors.Chapter 2.6 in Fundamentals of Electric CircuitsChapter 6.7 in Electric Circuit Fundamentals
Work through examples include a series-parallel resistor network (Example 4).Chapter 7.2 in Fundamentals of Electric Circuits
Voltage DividersResistors in series share the same current
Vin
Voltage DividersResistors in series share the same current
From Kirchoff’s Voltage Law and Ohm’s Law :
22
11
210
IRV
IRV
VVVin
+
V1-
+V2_
Vin
Voltage DividersResistors in series share the same current
From Kirchoff’s Voltage Law and Ohm’s Law :
in
in
in
VRRRV
VRRRV
RRVV
IRV
IRV
VVV
0
2122
2111
2112
22
11
21
+
V1-
+V2_
Vin
Voltage DivisionThe voltage associated with one resistor Rn in a chain of multiple resistors in series is:
or
where Vtotal is the total of the voltages applied across the resistors.
totaleq
nn V
R
RV
totalS
ss
nn V
R
RV
1
Voltage DivisionThe percentage of the total voltage
associated with a particular resistor is equal to the percentage that that resistor contributed to the equivalent resistance, Req.The largest value resistor has the largest
voltage.
Example 1Find the V1, the voltage
across R1, and V2, the voltage across R2.
+
V1-
+V2_
Example 1Voltage across R1 is:
Voltage across R2 is:
Check: V1 + V2 should equal Vtotal
+
V1-
+V2_
tVV
tVkkkV
VRRRV
tVV
tVkkkV
VRRRV
total
total
377sin4.11
377sin20 434
377sin57.8
377sin20 433
2
2
2122
1
1
2111
8.57 sin(377t)V + 11.4 sin(377t) = 20 sin(377t) V
+V1 -
+V2 -
+V3 -
Example 2Find the voltages listed in the
circuit to the right.
Check: V1 + V2 + V3 = 1V
VV
VV
VV
VV
VV
VV
R
R
eq
eq
143.0
1 700/100
571.0
1 700/400
286.0
1 700/200
700
100400200
3
3
2
2
1
1
+V1 -
+V2 -
+V3 -
Symbol for Parallel ResistorsTo make writing equations simpler, we use a symbol to indicate that a certain set of resistors are in parallel.
Here, we would write
R1║R2║R3
to show that R1 is in parallel with R2 and R3. This also means that we should use the equation for equivalent resistance if this symbol is included in a mathematical equation.
Current DivisionAll resistors in parallel share the same voltage
+
Vin
_
Current DivisionAll resistors in parallel share the same voltage
From Kirchoff’s Current Law and Ohm’s Law :
33
22
11
3210
RIV
RIV
RIV
IIII
in
in
in
in
+
Vin
_
Current DivisionAll resistors in parallel share the same voltage
+
Vin
_
in
in
in
IRRR
RRI
IRRR
RRI
IRRR
RRI
213
213
312
312
321
321
Current DivisionAlternatively, you can reduce the number of resistors in parallel from 3 to 2 using an equivalent resistor.
If you want to solve for current I1, then find an equivalent resistor for R2 in parallel with R3.
+
Vin
_
Current Division
+
Vin
_
ineq
eqeq I
RR
RI
RR
RRRRR
11
32
3232 and where
Current DivisionThe current associated with one resistor R1 in parallel with one other resistor is:
The current associated with one resistor Rm in parallel with two or more resistors is:
totalm
eqm I
R
RI
totalI
RR
RI
21
21
where Itotal is the total of the currents entering the node shared by the resistors in parallel.
Current DivisionThe largest value resistor has the smallest
amount of current flowing through it.
Example 3Find currents I1, I2, and I3 in the circuit to the
right.
Example 3 (con’t)
Check: I1 + I2 + I3 = Iin
AI
AI
AI
AI
AI
AI
Req
727.0
4 600/109
09.1
4 400/109
18.2
4 200/109
109600140012001
3
3
2
2
1
1
1
Example 4The circuit to the
right has a series and parallel combination of resistors plus two voltage sources.Find V1 and VpFind I1, I2, and I3
+
V1
_
+
Vp
_
I1
I2 I3
Example 4 (con’t)First, calculate
the total voltage applied to the network of resistors. This is the addition
of two voltage sources in series.
+
V1
_
+
Vp
_
I1
I2 I3
+
Vtotal
_
)20sin(5.01 tVVVtotal
Example 4 (con’t)Second, calculate
the equivalent resistor that can be used to replace the parallel combination of R2 and R3.
+
V1
_
+
Vp
_
I1
+
Vtotal
_
80100400
100400
1
1
32
321
eq
eq
eq
R
R
RR
RRR
Example 4 (con’t)To calculate the
value for I1, replace the series combination of R1 and Req1 with another equivalent resistor.
I1
+
Vtotal
_
280
80200
2
2
112
eq
eq
eqeq
R
R
RRR
Example 4 (con’t)I1
+
Vtotal
_
)20sin(79.157.3280
)20sin(5.0
280
1280
)20sin(5.01
1
1
1
21
tmAmAI
tVVI
tVVI
R
VI
eq
total
Example 4 (con’t)To calculate V1,
use one of the previous simplified circuits where R1 is in series with Req1.
+
V1
_
+
Vp
_
I1
+
Vtotal
_
)20sin(357.0714.0
or
1
111
1
11
tVVV
IRV
VRR
RV total
eq
Example 4 (con’t)To calculate Vp:
Note: rounding errors can occur. It is best to carry the calculations out to 5 or 6 significant figures and then reduce this to 3 significant figures when writing the final answer.
+
V1
_
+
Vp
_
I1
+
Vtotal
_
)20sin(143.0287.0
or
or
1
11
11
1
tVVV
VVV
IRV
VRR
RV
p
totalp
eqp
totaleq
eqp
Example 4 (con’t)Finally, use the
original circuit to find I2 and I3. +
V1
_
+
Vp
_
I1
I2 I3
)20sin(357.0714.0
or
2
12
12
132
32
tmAmAI
IR
RI
IRR
RI
eq
Example 4 (con’t)Lastly, the
calculation for I3.+
V1
_
+
Vp
_
I1
I2 I3
)20sin(43.186.2
or
or
3
213
13
13
132
23
tmAmAI
III
IR
RI
IRR
RI
eq
SummaryThe equations used
to calculate the voltage across a specific resistor Rn in a set of resistors in series are:
The equations used to calculate the current flowing through a specific resistor Rm in a set of resistors in parallel are:
totaln
eqn
totaleq
nn
VG
GV
VR
RV
total
total
I
I
eq
mm
m
eqm
G
GI
R
RI
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