1. LECTURE 14
Pick up : lecture notes

2. Kirchhoff’s rules for electric circuits
Typical problem
Given:
resistors,
batteries,
connections
Find:
current through,
voltage across
any circuit element
Gustav Robert Kirchhoff
Rule for “nodes”
“current in=current out”
Example: I1 J1 I2 I3 J2
Charge on the node
is conserved

3. II. Rule for “loops”
“Algebraic sum of potential changes = zero” E1 E2
Energy is
conserved

4. “Loop rule”: Sum of E’s = sum of IR’s
II. Rule for “loops”(cont’d)
“Algebraic sum of potential changes = zero”
Alternate form:
“Loop rule”: Sum of E’s = sum of IR’s

5. Basic strategy: WATCH FOR +/- SIGNS!!
Label each battery with + (long side) and – (short side)
Label the current in each branch with a symbol and an arrow (arbitrary direction).
Apply Kirchhoff’s node rule to one node
Apply Kirchhoff’s loop rule to a loop:
Choose a direction (clockwise or counter-clockwise) to “go” around a loop
Battery E is positive, if your direction of circulation takes you from the “-” pole to the “+” pole of the battery
Product IR is positive if direction of circulation coincides with current direction on the resistor
Write “Sum of E’s = Sum of IR’s”
Repeat 3-4 until # eqs. = # unknowns
Solve for unknowns (currents, for e.g.)

6. Problem 1: Calculate the current in the circuit
R1= 5 kW
R2= 15 kW I
E=12 V

7. CONCEPT TEST:
Two identical light bulbs A and B are wired in
series and connected to a power supply, as shown
in the figure. When the switch is closed, the
brightness of bulb A
1. increases
2. remains the same
3. decreases A B

8. Problem 2: Calculate the current in each branch of
the circuit below:
R1= 5 kW I1 I2
R2= 5 kW
R3= 15 kW I
+ -
E=12 V
Kirchhoff’s rule for
node 1
“blue” loop
“green” loop

9. CONCEPT TEST:
The light bulbs in the circuit are identical. When the switch is closed
1. both go out
2. A becomes brighter
3. B becomes brighter
4. A becomes dimmer
5. B becomes dimmer
6. combination of 1-5
7. nothing changes
12 V
12 V
12 V