Ch 31 Physics for Scientists and Engineers

Ch 31 Physics for Scientists and Engineers - FUNDAMENTALS...

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FUNDAMENTALS OF CIRCUITS 31.1. Solve: of lead from a mechanical pencil is From Table 30.1, the resistivity of carbon is p = 3.5 x R m. From Equation 31.3, the resistance p~ p~ (3.5 x lo-' R m)(0.06 m) A m' n(0.35~10-' m)' R=-=-= = 5.5 R 31.2. Solve: length L of a wire with a cross-sectional area A and having a resistance R is (a) From Table 30.1, the resistivity of aluminum is p = 2.8 x lo-' R m. From Equation 31.3, the (IO x lo4 m)z(lOOO R) L=-= = 3.57 m 2.8 x lo4 R m P (b) The number of turns is the length of the wire divided by the circumference of one turn. Thus, 3.57 m 2n(1.5 x 10" m) = 379 31.3. Solve: The current I in a wire when a potential difference is applied to the ends of the wire can be obtained from Ohm's law: I = AV/R. Using R = pL/A, (3.0 V)lc(0.40 x m)' (1.5~ lo4 R m)(0.50 m) = 2.0 A where we have used nichrome's resistivity from Table 30.1. 31.4. Solve: The potential difference between the ends of a copper wire that carries a current can be obtained from Ohm's law and the relationship R = p L/A . Finding the resistivity of copper from Table 30.1, L A ~(0.5 x m)' (3.0 A)(1.7 x lo-* C2 m)(0.20 m) AV=fR=Ip-= = 13 mV 31.5. Solve: diameter copper wire. That is, We need an aluminum wire whose resistance and length are the same as that of a 0.50-mm- =9 r. = ErCu = 2.8 x \1-(0.25 R m mm) = 0.32 mm 1.7 x IO-' R m We need a 0.64-mm-diameter aluminum wire. 31.6. Solve: The slope of the I versus AV graph, according to Ohm's law, is the resistance R. From the graph in Figure Ex3 I .6, the slope of the material is 31-1 ~.-x. ..._. ..I . "... ._,
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31-2 Chapter 31 31.7. Solve: 9vf - i;. . From the circuit in Figure Ex37.1, we see that 50 R and 100 R resistors are connected in series across the battery. Another resistor of 75 R is also connected across the battery. 31.8. Solve: that resistor is connected to resistor and a capacitor in parallel. In Figure Ex31.8, the positive terminal of the battery is connected to a resistor. The other end of 31.9. Model: Assume that the connecting wires are ideal. Visualize: Solve: I, = 10 V/5 SZ = 2 A downward. Using Kirchhoff's junction law, we see that Please refer to Figure Ex3 1.9. The current in the 2 R resistor is = 6 V/2 R = 3 A to the left. The current in the 5 R resistor is I = i- I2 = 3 A -I- 2 A = 5 A This current flows toward the junction, that is, downward. 31.10. Model: The batteries and the connecting wires are ideal. Visualize: Solve: (a) Choose the current I to be in the clockwise direction. If I ends up being a positive number, then the current really does flow in this direction. If I is negative, the current really flows counterclockwise. There are no junctions, so I is the same for all elements in the circuit. With the 9 V battery being labeled 1 and the 6 V battery being labeled 2. Kirchhoff s loop law is XAY =A%at, +AVK +4V,,, =+&, -lR-&2 =O Please refer to Figure Ex3 1.10. Note the signs: Potential is gained in battery I, but potential is lost both in the resistor and in battery 2. Because I is positive, we can say that I = 0.100 A flows from left to right through the resistor.
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Ch 31 Physics for Scientists and Engineers - FUNDAMENTALS...

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