Lecture 14
Kirchhoff’s laws.
ACT: 2 resistors, 2 batteries
In the circuit below, the switch is initially open. When
the switch is closed, the current
through the bottom resistor:
9 V
R
A. Increases
B. Decreases
C. Stays the same
9 V
9 V
R
ACT: 2 resistors, 2 batteries
In the circuit below, the switch is initially open. When
the switch is closed, the current
through the bottom resistor:
9 V
R
A. Increases
B. Decreases
C. Stays the same
DEMO:
Batteries in series
and parallel
9 V
9 V
R
When the switch is open, the potential difference
across both resistors is 18 V. Since they are equal, it
must be 9 V across each.
Adding the third battery does
not change the potential
difference across the
resistor, it is still 9 V!
9 V
R
9 V
9 V
R
Kirchhoff’s laws
Junction rule or KCL (Kirchhoff’s current law):
The algebraic sum of currents entering a junction
must be equal to the algebraic sum of currents leaving
the junction.
Loop rule or KVL (Kirchhoff’s voltage law):
The algebraic sum of changes in potential in any closed
circuit loop must equal zero.
EXAMPLE: Two-loop circuit
Determine the currents through the elements of
this circuit.
ε2
R1
ε1
R2
R3
Step 1: How many distinct currents are there? 3
We will need 3
equations
Draw them!
ε2
R1
ε1
R2
I1
I2
R3
I3
Step 2: How many junctions?
Write N - 1
junction equations
N = 2
I2 + I3 = I1
ε2
R1
ε1
R2
I1
I2
R3
I3
Step 3: Draw as many loops as you need to complete
the number of required equations (found in step 1)
Many possible loops. In this case we only need
2 of them.
ε2
R1
ε1
R2
I1
I2
R3
I3
Step 4: Write the loop equations for each chosen loop
ε1 – I1 R1 – I2 R2 = 0
ε2
R1
ε1
R2
I1
I2
R3
I3
ε2 – I2 R2 + I3 R3 = 0
Path and current in
opposite direction
ε2
R1
ε1
R2
I1
I2
R3
I3
We only need two equations, but let us do a third one just for fun…
ε1 – I1 R1 - ε2 – I3 R3 = 0
Path is “against” the battery (moving from – to + ends)
ε2
R1
ε1
R2
I1
I2
R3
I3
Solve the system: 3 eqns, 3 unknowns
I2 + I3 = I1
ε1 – I1R1 – I2R2 = 0
ε2 – I2R2 + I3R3 = 0
ε1 = 12 V
ε2 = 24 V
For R = 5 Ω
1
R2 = 3 Ω
I1 = 0.25 A
I2 = 3.6 A
I3 = –3.3 A
R3 = 4 Ω
Nice circuit animations (very visual):
I3 flows
opposite to our
assumption
http://phet.colorado.edu/web-pages/simulations-base.html