Download Kirchhoff’s Law

11.6
p.519-522
KIRCHHOFF’S LAW
Kirchhoff’s Law VOLTAGE
 KVL: in any complete path in
an electric circuit, the total
electric potential increase at
the source (s) is equal to the
total electric potential
decrease throughout the rest
of the circuit
 RECALL:
 Placing all the components in a circuit along one
path is called placing them in series.
 A series circuit has only one path for the charges
to follow.
KVL in series circuits
 Has only one complete path, so loads must SHARE
the amount of electric potential
 The total electric potential across the loads MUST
add up to the electric potential at the source
 Vseries= V1 + V2 + V3 + Vn …
 RECALL:
 A parallel circuit is a closed circuit in which the
current has more than one path, or branch, to
follow.
 The point at which the path splits is called a
junction.
KVL in parallel circuits
 Has more than one complete
path, so electric potential
decrease across each load must
be the same as the potential
increase at the source
 Vparallel= V1= V2= V3 = Vn …
Kirchhoff’s Current Law
 KCL: in a closed circuit, the
amount of current entering a
junction is equal to the amount
of current exiting a junction.
 *junctions are points where the
current can split to follow more
than one path
KCL Series circuit
 A series circuit has only one
path. So there can only be one
possible current. If you know
the current at source you
know it everywhere in circuit
Iseries=I1=I2=I3= In …
KCL parallel circuits
 A parallel circuit has more than one
complete path, so the current can split,
depending on the number of paths. The
more complete paths there are, the more
ways current can be divided among the
paths.
 Iparallel= I1 + I2 + I3 +…In
Mixed Circuits
 Almost all electric devices
contain a combination of
series and parallel circuits.
 These circuits are called
mixed circuits.
p. 521 KVL mixed circuit
 You can look at this circuit
as two complete circuits
 The red path from the
source to lamp 1, lamp
2, lamp 3, and back to
the source is one path
 Vtotal= V1 + V2 +V3
 The blue path from the
source to lamp 1, lamp
2, lamp 4 and back to
source is another path
 Vtotal= V1 + V2 + V4
p. 521 KCL to mixed circuit
 Using the same circuit you can find the current. The
current entering a junction is equal to the current
exiting the junction
 The current in a series circuit is constant, and in a
parallel circuit it splits
 Iseries=I1=I2 and Iparallel=I3 + I4
Kirchhoff’s Current Law in Mixed Circuit
Section 11.6
In this circuit, the current through lamp 2 is 0.75 A, and the current through lamp
4 is 0.15 A. Determine the current through each of the other lamps. Use the data
from the problem to fill in the Given information. Then, use the variables to fill in
the Analysis.
Given: I2 =
I4 =
Required: I1; I3
Analysis: Iseries =
Iparallel =
=
+
Kirchhoff’s Current Law in Mixed Circuit
Section 11.6
In this circuit, the current through lamp 2 is 0.75 A, and the current through lamp
4 is 0.15 A. Determine the current through each of the other lamps. Use the data
from the problem to fill in the Given information. Then, use the variables to fill in
the Analysis.
Given: I2 =
0.15 A
0.75 A I4 =
Required: I1; I3
Analysis: Iseries =
Iparallel =
I1 =
I2
I3+
I4
Section 11.6
Kirchhoff’s Current Law in Mixed Circuit
In this circuit, the current through lamp 2 is 0.75 A,
and the current through lamp 4 is 0.15 A.
Determine the current through each of the other
lamps. Use the data from the problem to complete
the calculation. Then, complete the final
statement. Click the screen for a hint.
Iseries = Iparallel
Solution:
Iseries =
=
Iseries =
=
Iparallel
I1
I2
I3
=
I4
+
=
+
=
Statement: The current through lamp 1 is __________. The current through lamp
3 is __________.
Section 11.6
Kirchhoff’s Current Law in Mixed Circuit
In this circuit, the current through lamp 2 is 0.75 A,
and the current through lamp 4 is 0.15 A.
Determine the current through each of the other
lamps. Use the data from the problem to complete
the calculation. Then, complete the final
statement. Click the screen for a hint.
Iseries = Iparallel
Solution:
Iseries =
I1 =
I2
Iseries =
I1 =
0.75 A
Iparallel
I1
I2
I3
I3 +
=
0.75 A
=
I3
0.60 A
=
I3
I4
I4
+
0.15 A
0.75 A The current through lamp
Statement: The current through lamp 1 is __________.
0.60 A
3 is __________.
Homework
 P. 522 practice #1-2
 P. 522 questions #1-2
p. 522 #1 figure 7
a) Find V2, V4, and V5 if Vsource=60.0V,
V1=20.0V, and V3=15V
b) Find Isource, I3, and I4 if I1=0.70A,
I3=0.10A, and I5=0.20A
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1 - Reduce the circuit to a simple series
circuit by using equivalent resistors.
2 - Determine the total resistance and total
current of the series circuit using the
equation for RT for series circuits and Ohm’s
law.
3 - Determine the voltage drop across each
resistor in the circuit using Ohm’s law.
4 - Redraw the original circuit with the
voltage drops beside each resistor.
Remember that the voltage drop across the
parallel resistors will be the same.
5 - Determine the current through the
parallel resistors using Ohm’s law.