Physics Assignment

# Physics Assignment - Phys 122 Introductory Physics II...

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Unformatted text preview: Phys 122 Introductory Physics II TUTORIAL 4 Electric Circuits - Kirchhoff’s Rules. Key 1. Apply Kirchhoff's rules to find the currents in the circuit shown. Label the currents and choose the current directions. It is not necessary to "guess" the true current directions. Let I 1 be up through the 12 V battery, I 2 be down through the 15 Ω resistor, and I 3 be down through the 25 Ω resistor. Apply the loop rule (drops = rises) to the left loop starting at the upper left corner and proceeding clockwise. I 3 (25 Ω ) = 12 V, then I 3 = 0.48 A . Repeat for the outside loop I 2 (15 Ω ) = 6.0 V + 12 V, then I 2 = 1.2 A. Apply the junction rule (currents in = currents out) at the top junction. I 1 = I 2 + I 3 which gives I 1 = 1.7 A . Note that all of the currents are positive, indicating that we "guessed" their directions correctly. 2. Consider the circuit in the drawing. Determine (a) the magnitude of the current in the circuit and (b) the magnitude of the voltage between the points labeled A and B . (c) State which point, A or B , is at the higher potential. 1 a. We assume that the current is directed clockwise around the circuit. Starting at the upper-left corner and going clockwise around the circuit, we set the potential drops equal to the potential rises: (5.0 Ω ) I + (27 Ω ) I + 10.0 V + (12 Ω ) I + (8.0 Ω ) I Potential drops 1 2 4 4 4 4 4 4 4 4 4 3 4 4 4 4 4 4 4 4 4 = 30.0 V Potential rises 1 2 3 Solving for the current gives I = 0.38 A ....
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## This note was uploaded on 04/18/2008 for the course PHYSICS 123 taught by Professor Wrzesneski during the Winter '08 term at UBC.

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Physics Assignment - Phys 122 Introductory Physics II...

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