LECTURE 14 Pick up : lecture notes

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

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

“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

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.)

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

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

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

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