Transcript Voltage Current Dividers

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
 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 Circuits
 Chapter 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 Dividers
Resistors in series share the same current
Vin
Voltage Dividers
Resistors in series share the same current
From Kirchoff’s Voltage Law and Ohm’s Law :
+
V1
Vin
0  Vin  V1  V2
V1  IR1
V2  IR2
+
V2
_
Voltage Dividers
Resistors in series share the same current
From Kirchoff’s Voltage Law and Ohm’s Law :
+
0  Vin  V1  V2
V1
-
V1  IR1
Vin
V2  IR2
+
V2
_
V2  V1 R1 R2
V1  R1 R1  R2 Vin
V2  R2 R1  R2 Vin
Voltage Division
The voltage associated with one resistor Rn in a chain
of multiple resistors in series is:


 R 
Vn   S n  Vtotal
 R 
s
 
s 1

or
 Rn 
Vn  
 Vtotal
 Req 
where Vtotal is the total of the voltages applied across
the resistors.
Voltage Division
 The 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 1
 Find the V1, the voltage across R1,
and V2, the voltage across R2.
+
V1
-
+
V2
_
Example 1
 Voltage across R1 is:
V1  R1 R1  R2 Vtotal
V1  3k 3k  4k  20V sin 377t 
V1  8.57V sin 377t 
+
V1
-
 Voltage across R2 is:
V2  R2 R1  R2 Vtotal
V2  4k 3k  4k  20V sin 377t 
V2  11.4V sin 377t 
 Check: V1 + V2 should equal Vtotal
8.57 sin(377t)V + 11.4 sin(377t) = 20 sin(377t) V
+
V2
_
Example 2
 Find the voltages listed in the
circuit to the right.
+
V1
-
+
V2
-
+
V3
-
Example 2 (con’t)
Req  200  400  100
Req  700
V1  200 / 700 1V 
V1  0.286V
V2  400 / 700 1V 
+
V1
-
+
V2
-
V2  0.571V
V3  100 / 700 1V 
V3  0.143V
Check: V1 + V2 + V3 = 1V
+
V3
-
Symbol for Parallel Resistors
To 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 Division
All resistors in parallel share the same voltage
+
Vin
_
Current Division
All resistors in parallel share the same voltage
From Kirchoff’s Current Law and Ohm’s Law :
+
0   I in  I1  I 2  I 3
Vin
Vin  I1 R1
_
Vin  I 3 R3
Vin  I 2 R2
Current Division
All resistors in parallel share the same voltage
+
Vin
I1 
I2 
_
I3 
R2 R3
R1  R2 R3
R1 R3
R2  R1 R3
R1 R2
R3  R1 R2
I in
I in
I in
Current Division
Alternatively, you can reduce the number
of resistors in parallel from 3 to 2 using an
equivalent resistor.
+
Vin
_
If you want to solve for current I1, then
find an equivalent resistor for R2 in
parallel with R3.
Current Division
+
Vin
_
Req
R2 R3
where Req  R2 R3 
and I1 
I in
R2  R3
R1  Req
Current Division
The current associated
with one resistor R1 in
parallel with one other
resistor is:
 R2 
I1  
 I total
 R1  R2 
 The current associated
with one resistor Rm in
parallel with two or more
resistors is:
 Req 
I m    I total
 Rm 
where Itotal is the total of the currents entering the
node shared by the resistors in parallel.
Current Division
 The largest value resistor has the smallest amount of
current flowing through it.
Example 3
Find currents I1, I2, and I3 in the circuit to the right.
Example 3 (con’t)
Req  1 200   1 400   1 600   109
1
I1  109 / 200 4 A
I1  2.18A
I 2  109 / 400 4 A
I 2  1.09A
I 3  109 / 600 4 A
I 3  0.727A
Check: I1 + I2 + I3 = Iin
Example 4
 The circuit to the
I1
right has a series and
parallel combination
of resistors plus two
voltage sources.
+
V1
_
 Find V1 and Vp
 Find I1, I2, and I3
I2
I3
+
Vp
_
Example 4 (con’t)
I1
 First, calculate the
total voltage applied
to the network of
resistors.
+
+
V1
 This is the addition of
_
two voltage sources in
series.
Vtotal  1V  0.5V sin(20t )
Vtotal
I2
I3
+
Vp
_
_
Example 4 (con’t)
I1
 Second, calculate the
equivalent resistor
that can be used to
replace the parallel
combination of R2
and R3.
Req1 
R2 R3
R2  R3
400100 
Req1 
400  100
Req1  80
+
+
V1
_
Vtotal
+
Vp
_
_
Example 4 (con’t)
I1
 To calculate the value
for I1, replace the
series combination of
R1 and Req1 with
another equivalent
resistor.
Req 2  R1  Req1
+
Vtotal
Req 2  200  80
Req 2  280
_
Example 4 (con’t)
Vtotal
I1 
Req 2
1V  0.5V sin(20t )
I1 
280
1V
0.5V sin(20t )
I1 

280
280
I1  3.57m A 1.79m Asin(20t )
I1
+
Vtotal
_
Example 4 (con’t)
I1
 To calculate V1, use
one of the previous
simplified circuits
where R1 is in series
with Req1.
R1
V1 
Vtotal
R1  Req
+
+
V1
_
Vtotal
or
+
V1  R1 I1
Vp
V1  0.714V  0.357V sin(20t )
_
_
Example 4 (con’t)
 To calculate Vp:
Vp 
Req1
R1  Req1
Vtotal
I1
+
+
or
V p  Req1 I1
or
V p  Vtotal  V1
_
Vtotal
V p  0.287V  0.143V sin(20t )
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
_
_
Example 4 (con’t)
 Finally, use the
I1
original circuit to
find I2 and I3.
+
V1
R3
I2 
I1
R2  R3
I2
or
I2 
_
Req1
R2
I1
I 2  0.714m A 0.357m Asin(20t )
I3
+
Vp
_
Example 4 (con’t)
 Lastly, the
I1
calculation for I3.
+
R2
I3 
I1
R2  R3
V1
or
_
I3 
Req1
R3
I1
or
I 3  I1  I 2
I 3  2.86m A 1.43m Asin(20t )
I2
I3
+
Vp
_
Summary
 The equations used to
calculate the voltage
across a specific resistor
Rn in a set of resistors in
series are:
 Rn 
Vn  
 Vtotal
 Req 
 Geq 
Vn  
 Vtotal
 Gn 
 The equations used to
calculate the current
flowing through a specific
resistor Rm in a set of
resistors in parallel are:
Im 
Req
Rm
I total
Gm
Im 
I total
Geq