20.8 Circuits Wired Partially in Series and Partially in Parallel
Example: In the given circuit:
(a) How much power is drawn from the battery?
(b) How much current flows through each resistor? And in what direction?
In examining a circuit,
you start by trying to find
combinations of resistors
that are in series and/or
in parallel
1
20.8 Circuits Wired Partially in Series and Partially in Parallel
220Ω and 250Ω in series:
RS = 220Ω + 250Ω = 470Ω
180Ω and 470Ω in parallel:
1
1
1
=
+
= 7.683 × 10 −3 Ω −1
RP 180Ω 470Ω
RP =
1
= 130Ω
−3
−1
7.683 × 10 Ω
I
110Ω and 130Ω in series:
RS = 110Ω + 130Ω = 240Ω
V 2 ( 24V ) 2
(a) P = IV =
=
= 2.4W
R
240Ω
2
20.8 Circuits Wired Partially in Series and Partially in Parallel
I
V
24V
=
= 0.10 A
R
240Ω
I 110 Ω = 0.10 A (left to right)
I=
I
I
V110 Ω = I 100 Ω (110 Ω)
= (0.10 A )(110 Ω) = 11 V
VAB = VB − VA = − I 130 Ω (130 Ω)
= −(0.10 A)(130 Ω) = −13 V
0.10 A
I 470 Ω =
0.10 A
0.10 A
0.072 A
0.072 A
=
VAB
=
13 V
180 Ω 180 Ω
= 0.072 A (down)
I 180 Ω =
I 180 Ω
VAB
470 Ω
13 V
= 0.028 A
470 Ω
0.028 A
I 220 Ω = I 250 Ω
= 0.028 A (down)
3
20.10 Kirchhoff’s Rules
In many circuits, especially ones involving loops
and multiple voltage supplies:
The resistors cannot be combined into series and
parallel equivalents.
For example, in the circuit to the left, the three
resistors are clearly not in series, since they do not
share the same current
Here I2=I1+I3 by the junction rule
3.0 V
They are also NOT in parallel
The presence of the two batteries with different
voltage makes it so that the voltage across the
three resistors are NOT the same.
We need a more general approach to solving battery-resistor networks:
Kirchoff’s Rules:
Based on two basic principles
(a) Conservation of charge
(b) The potential well-defined and measuable (i.e. single-valued) at any point
We also include the basic properties of the circuit elements
4
20.10 Kirchhoff’s Rules
KIRCHHOFF’S RULES
(a) Junction Rule
7A = 2A + 5A
The total current directed into a junction must
equal the total current directed out of the junction.
Important Point: For each continuous segment
of the circuit (between branches), one should
label a current with a presumed direction. This
DOES NOT need to be the actual direction.
After solving the circuit, some of these current will
solve to positive values—these are flowing in the
presumed directions
I1+I4 = I2+I3+I5
I1=I2+I3
The currents that solve to negative values flow in
the direction opposite of the presumed labeling.
An analogy can be drawn from the flow of water
in channels. The junctions in a circuit is
equivalent to junctions in water ways.
5
20.10 Kirchhoff’s Rules
(b) Loop rule: Around any closed circuit loop, the sum of the potential drops
equals the sum of the potential rises.
The loop rule expresses conservation of energy in terms of the electric
potential and states that for a closed circuit loop, the total of all potential
rises is the same as the total of all potential drops.
Here we know the equivalent series
resistance is RS=R1+R2=6Ω
So I = V/R =12V / 6Ω = 2A
In practice you go around a loop once
sequentially
Starting at a point on the circuit that is
between circuit elements
Here we start at S
There are no junctions: there is only one
current: We label the current counter-clockwise (in the direction of the battery).
S
Traversing the circuit clock-wise:
(a) Voltage drop along the current over 5Ω:
∆V1 = −IR = −(2A)(5Ω) = −10V
(b) Voltage drop along the current over 1Ω:
∆V2 = −IR = −(2A)(1Ω) = −2V
(c) Voltage drop from – to + side of a battery:
∆V3 = +E = 12V
Loop complete !
Note ∆V1 + ∆V2 + ∆V3 =0
6
20.10 Kirchhoff’s Rules
Voltage rise or drop through a resistor
or battery. In each case we are
traversing from point a to b
There are 4 cases:
Resistor: along
presumed current
If we were solving this circuit, we’d take
the following steps
(a) Label the current(s)
(b) Eliminate some of the current(s) by
enforcing junction rules
(c) Determine how many loops we need:
N = number of remaining unknown,
independent currents
(d) Traverse through N different
(independent) loops and write down
the loop rule for each  gives N
linear equations in the N unknown
currents
Resistor: against
presumed current
Voltage source:
From – to +
terminal
Voltage source:
From + to –
terminal
7
20.10 Kirchhoff’s Rules
Example: Using Kirchhoff’s Loop Rule,
determine the current in the circuit.
Again, this is a one loop circuit, no
junctions and branches
We have single unknown current I,
which we presume to be counterclock-wise (CCW). This is an
educated guess, because the 24V
battery should push the current (of
fictitious positive charge carriers) more
CCW than the 6V battery tries to push it
CW.
Going through the loop starting at
bottom left corner (point S)
− I (12 Ω) − 6.0 V − I (8.0 Ω) + 24 V = 0
Rearranging :
− I ( 20 Ω) + 18 V = 0
→ ( 20 Ω) I = 18 V
18 V
I=
= 0.90 A (clockwise)
20 Ω
8
20.10 Kirchhoff’s Rules
Example: Using Kirchhoff’s Rules, determine
the power consumed by each of the 3 resistors
The circuit has 3 branches and 2 junctions
We label 3 currents as shown
The presumed directions are motivated by
the fact that the 5V battery will push more
current right to left than the 3V in the
opposite direction
Apply Junction Rule:
A
B
Junction A : I IN = I 2 , I OUT = I 1 + I 3
→ I 2 = I1 + I 3
Junction B : I IN = I 1 + I 3 , I OUT = I 2
→ I1 + I 3 = I 2
3.0 V
Notice these constraints are the SAME
There will generally be one more junction than constraints on the currents
(this is because we are dealing with closed circuit loops
Now we choose to eliminate one current based on the one constraint:
This is an arbitrary choice. We have to find the power consumed each resistor  we
have to find all three currents. We will eliminate I3 because it goes through the fewest
components. If for example you are only asked for the power in one of the 3 resistors
then you should eliminate first a different current (leads to faster solution).
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20.10 Kirchhoff’s Rules
I 2 = I1 + I 3 → I 3 = I 2 − I1
We now apply Loop Rule
There are more loops (3) than we need (2) for
solving for the two independent currents:
One outer loop and two inner loops
I2-I1
Loop 2
I2
A
B
I1
Loop 1
There are generally more loops than we need.
We need to choose a combination of 2 loops that
will include EVERY COMPONENT
I1
We pick the two inner loops as shown
3.0 V
Loop 1 : traverse in counter - clock - wise direction starting at B
+ 5 V − I 2 (1.0 Ω) − I 1 (1.0 Ω) − 3.0 V = 0
(1)
Loop 2 : traverse in clock - wise direction starting at B
+ 5 V − I 2 (1.0 Ω) − ( I 2 − I 1 )( 2.0 Ω) = 0
(2)
Rearranging (1) and (2) into the standard form,
and dropping SI units (where then the currents are solved in amps) :
I1 + I 2 = 2
(1) →
(1' )
( 2) →
− 2 I 1 + 3I 2 = 5
(2' )
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20.10 Kirchhoff’s Rules
I1 + I 2 = 2
(1' )
− 2 I 1 + 3I 2 = 5
(2' )
I2-I1
This is a 2x2 system of linear equations.
Loop 2
If the problems asks for only, for example, the power
consumed in the resistor on the 3V battery branch…
Then one should solve for I1 first.
I2
A
Here we are free to choose: let us eliminate I1 first.
There are a few ways to do this: the one that can be
applied to larger (say 3x3) systems:
B
I1
Loop 1
I1
Operate on equations: 2×(1’) + (2’)
2 I1 + 2 I 2 = 4
2 × (1' )
− 2 I 1 + 3I 2 = 5
(2' )
0 I 1 + 5 I 2 = 9 → (5.0 Ω) I 2 = 9.0 V
3.0 V
It is useful to double check your results with a
loop that was not used: the outerloop, clockwise
starting at point B:
+ 3.0 V + I 1 (1.0 Ω) − I 3 ( 2.0 Ω)
9.0 V
I2 =
= 1.8 A
5.0 Ω
substituting into equation (1' )
(1.0 Ω) I 1 + (1.0 Ω) I 2 = 2.0 V
top : P2.0 Ω = (I 3 ) ( 2.0 Ω) = (1.6 A ) 2 ( 2.0 Ω) = 5.1 W
→ I 1 = 2.0 A - 1.8 A = 0.20 A
mid : P1.0 Ω = (I 2 ) (1.0 Ω) = (1.8 A ) 2 (1.0 Ω) = 6.5 W
and I 3 = I 2 − I 1 = 1.8 A − 0.20 A = 1.6 A
= +3.0 V + (0.20 A)(1.0 Ω) − (1.6 A)( 2.0 Ω)
= +3.0 V + 0.20 V − 3.2 V = 0 (correct! )
2
2
bot : P1.0 Ω = (I 1 ) (1.0 Ω) = (0.20 A ) 2 (1.0 Ω) = 0.040 W
2
11
20.10 Kirchhoff’s Rules
I1 + I 2 = 2
(1' )
− 2 I 1 + 3I 2 = 5
(2' )
I2-I1
This is a 2x2 system of linear equations.
Loop 2
If the problems asks for only, for example, the power
consumed in the resistor on the 3V battery branch…
Then one should solve for I1 first.
I2
A
Here we are free to choose: let us eliminate I1 first.
There are a few ways to do this: the one that can be
applied to larger (say 3x3) systems:
B
I1
Loop 1
I1
Operate on equations: 2×(1’) + (2’)
2 I1 + 2 I 2 = 4
2 × (1' )
− 2 I 1 + 3I 2 = 5
(2' )
0 I 1 + 5 I 2 = 9 → (5.0 Ω) I 2 = 9.0 V
3.0 V
It is useful to double check your results with a
loop that was not used: the outerloop, clockwise
starting at point B:
+ 3.0 V + I 1 (1.0 Ω) − I 3 ( 2.0 Ω)
9.0 V
I2 =
= 1.8 A
5.0 Ω
substituting into equation (1' )
(1.0 Ω) I 1 + (1.0 Ω) I 2 = 2.0 V
top : P2.0 Ω = (I 3 ) ( 2.0 Ω) = (1.6 A ) 2 ( 2.0 Ω) = 5.1 W
→ I 1 = 2.0 A - 1.8 A = 0.20 A
mid : P1.0 Ω = (I 2 ) (1.0 Ω) = (1.8 A ) 2 (1.0 Ω) = 6.5 W
and I 3 = I 2 − I 1 = 1.8 A − 0.20 A = 1.6 A
= +3.0 V + (0.20 A)(1.0 Ω) − (1.6 A)( 2.0 Ω)
= +3.0 V + 0.20 V − 3.2 V = 0 (correct! )
2
2
bot : P1.0 Ω = (I 1 ) (1.0 Ω) = (0.20 A ) 2 (1.0 Ω) = 0.040 W
2
12