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Chapter 18 Kirchhoff’s Rules RC Circuits

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. 1) The sum of the currents entering a junction

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Chapter 18 Kirchhoff’s Rules RC Circuits

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

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

  3. Kirchhoff's Rules Junction 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)

  4. 5W I1 I I 3W 10W 2W I2 I3 2W V The Junction Rule says that I = I1 + I2 + I3

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

  6. 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.

  7. 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

  8. 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).

  9. I I2 R1 I1 R2 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 - I1R1 +I2R2

  10. V1 V2 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 + V2

  11. 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.

  12. 5W I1 I I 3W 10W 2W I2 I3 2W V This circuit has two junction points. Therefore, we can only use the junction rule ONE time.

  13. 12 W 1 W b e a I2 I0 6 W I1 12 V + 18 V _ d f c 1 W + _ Find the currents in each branch of this circuit. There are two junction points, so we can apply the junction ruleONCE. I0 = I1 + I2

  14. 12 W 1 W b e a 6 W 12 V + 18 V _ d f c 1 W + _ 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!)

  15. 12 W 1 W b e a I2 I0 6 W I1 12 V + 18 V _ d f c 1 W + _ Let’s go around bcda first. -(6W) I1 + 18 V - (12W) I0 = 0 (12W) I0 + (6W) I1 = 18 V

  16. 12 W 1 W b e a I2 I0 6 W I1 12 V + 18 V _ d f c 1 W + _ Let’s go around efcb second. 12 V - (1W) I2 + (6W) I1 - (1W) I2 = 0 (2W) I2 - (6W) I1 = 12 V

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