HW 15 - terry(ect328 – homework 15 – Turner –(59130 1...

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Unformatted text preview: terry (ect328) – homework 15 – Turner – (59130) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Note: Making use of the fact that the resistors and electric potentials are integer multiples of one another may make the solution less tedious. 247V 2 k Ω 4 k Ω 494V 8 k Ω 6 k Ω a b c d e What is the current flowing directly from “a” to “e”? Correct answer: 24 . 7 mA. Explanation: E 1 I 1 R 1 I 2 R 2 E 2 I ae I 4 R 4 I 3 R 3 a b c d e Let : R 1 = R = 2 k Ω , R 2 = 2 R = 4 k Ω , R 3 = 3 R = 6 k Ω , R 4 = 4 R = 8 k Ω , E 1 = E = 247 V , and E 2 = 2 E = 494 V . Note: Even though “a” and “e” are at the same electric potential (since they are connected by a wire), the wire can carry a current (since the wire has zero resistance). Basic Concepts: Resistors in Parallel and Series. Kirchhoff’s Laws summationdisplay V = 0 around a closed loop . summationdisplay I = 0 at a circuit node . Solution: For this, we simply apply Kirchhoff’s rules to the circuit. With the currents labeled as shown in the figure above, we get the loop equations E − I 1 R − I 4 R 4 = 0 2 E − I 2 R 2 − I 3 R 3 = 0 I 4 R 4 = I 3 R 3 , or E − I 1 R − I 4 (4 R ) = 0 (1) 2 E − I 2 (2 R ) − I 3 (3 R ) = 0 (2) I 4 (4 R ) = I 3 (3 R ) . (3) Solving these for the currents gives I 1 = E R − 4 I 4 (1 ′ ) I 2 = E R − 3 2 I 3 (2 ′ ) I 3 = 4 3 I 4 . (3 ′ ) We can also immediately get I 2 = E R − 2 I 4 . (2 ′′ ) Now we can use the node equation at node “c” I 1 + I 2 = I 3 + I 4 . Combining this with equations (1 ′ ), (2 ′′ ), and (3 ′ ) gives parenleftbigg E R − 4 I 4 parenrightbigg + parenleftbigg E R − 2 I 4 parenrightbigg = parenleftbigg 4 3 I 4 parenrightbigg + I 4 . Then I 4 = 6 25 E R Finally, using the node equation at node “a”, we have I ae = I 4 − I 1 = 1 5 E R = 1 5 247 V 2 k Ω · 1 kΩ 1000 Ω · 1000 mA 1 A = 24 . 7 mA . terry (ect328) – homework 15 – Turner – (59130) 2 Solution: E 1 I 1 R 1 I 2 R 2 E 2 I 34 R 34 a b c d e Figure 3 We can reduce the number of loops and node equations if we simplify the circuit by noticing that the resistors 4 R and 3 R are parallel, as in Fig. 3. Then 1 R 34 = 1 R 3 + 1 R 4 = R 3 + R 4 R 3 R 4 R 34 = R 3 R 4 R 3 + R 4 = (3 R ) (4 R ) 3 R + 4 R = 12 7 R . Applying Kirchhoff’s Laws to the two small loops and node “c” in this circuit, V − I 1 R − I 34 R 34 = 0 (4) 2 V − I 2 (2 R ) − I 34 R 34 = 0 (5) I 34 = I 1 + I 2 . (6) Subtracting equation (5) from equation (4) gives V − I 1 R − 2 V + I 2 (2 R ) = 0 . (7) Solving eqn. (7) for I 2 i gives I 2 = I 1 2 + V 2 R ....
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HW 15 - terry(ect328 – homework 15 – Turner –(59130 1...

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