Lecture 24 Chapter 18 Kirchhoff’s Rules RC Circuits Monday 22 March 1999 Physics 112

The Physics 112 Help Session Tuesdays 6:00 - 7:00 pm CANCELLED! Mondays 5:30 - 7:00 pm NSC Room 118

Kirchhoff's Rules Junction Rule Loop Rule 1) The sum of the currents entering a junction must equal the sum of currents leaving a junction. (Conservation of Charge) Loop Rule 2) The sum of potential differences across all the elements on any closed loop in a circuit must be zero. (Conservation of Energy)

2W 10W 5W 3W I I3 I2 I1 V The Junction Rule says that I = I1 + I2 + I3

2W 10W 5W 3W I I3 I2 I1 V We can use the loop rule multiple times: V - 3 I - 10 I2 - 2I = 0 V - 3I - 2I3 - 2I = 0 V - 3 I - 5 I1 - 2I = 0

We now have a system of 4 equations and 4 unknowns. We can therefore solve for each of the 4 unknown currents. V - 3 I - 10 I2 - 2I = 0 V - 3I - 2I3 - 2I = 0 V - 3 I - 5 I1 - 2I = 0 I = I1 + I2 + I3 Plug these three into the current equation.

10I = 8V - 40 I I = 0.16 V Plug this back into the other 3 equations... I1 = 0.04 V I2 = 0.02 V I3 = 0.1 V

For Applying Kirchhoff’s Rules Additional Notes For Applying Kirchhoff’s Rules A) assign a direction to the current in each branch of the circuit. Just GUESS!! If your guess is incorrect, the current will come out as a negative number, but the magnitude will still be correct! B) when applying the loop rule, you must choose a consistent direction in which to proceed around the loop (either clockwise or counterclockwise, your choice, but stick to it).

- I1R1 +I2R2 Still More Notes! 1) When you encounter a resistor in the direction of the current, the voltage drop is DV = - I R 2) When you encounter a resistor opposite the current, the voltage drop is DV = + I R R2 I R1 I1 I2 - I1R1 +I2R2

You Guessed It! 3) When you encounter an emf in the direction you’re going around the loop, the voltage change is +DV 4) When you encounter an emf opposite the the direction you’re going, the voltage change is -DV - V1 V1 V2 + V2

Last Page of Notes 5) The junction rule can only be applied n-1 times in a circuit with n junctions. 6) Each new equation you write must contain a current that you haven’t yet used. 7) To solve a system of equations with k unknown quantities, you need k independent equations.

2W 10W 5W 3W I I3 I2 I1 V This circuit has two junction points. Therefore, we can only use the junction rule ONE time.

I0 = I1 + I2 Find the currents in each branch of this circuit. 6 W 1 W 18 V 12 W 12 V + _ a f e c b d I2 I0 I1 Find the currents in each branch of this circuit. There are two junction points, so we can apply the junction rule ONCE. I0 = I1 + I2

3 How many possible loops are there? b c d a e f c d a b e f c b 18 V 12 W 12 V + _ a f e c b d 3 How many possible loops are there? b c d a e f c d a b e f c b We have 3 unknowns, so we need 3 equations total. Therefore, we need use only 2 of the 3 equations provided by the loop rule. (The junction rule gave us 1 equation already!)

Let’s go around bcda first. 6 W 1 W 18 V 12 W 12 V + _ a f e c b d I2 I0 I1 Let’s go around bcda first. -(6W) I1 + 18 V - (12W) I0 = 0 (12W) I0 + (6W) I1 = 18 V

Let’s go around efcb second. I0 6 W 1 W 18 V 12 W 12 V + _ a f e c b d I1 I2 Let’s go around efcb second. 12 V - (1W) I2 + (6W) I1 - (1W) I2 = 0 (2W) I2 - (6W) I1 = 12 V