Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

Transcript

Giambattista, Ch. 18 Problems: 1, 12, 22, 23, 40, 42, 44, 52, 55, 57, 60, 61 Note that in the printing of some of these equations, some symbols don’t always come through. “delta V” sometimes looks like a % sign. 1.Strategy Use the definition of electric current. (Current is defined as “charge per time”.) Solution Compute the total charge. 12. (a) Strategy Calculate the work done using Eq. (18-2) and the definition of electric current. Alternatively, you could think about Part b first, and multiply power by time to get energy. Solution Compute the amount of electrical energy supplied by the solar cell. (b) Strategy The power is equal to the rate at which the solar cell supplies electrical energy. Solution Find the average power by dividing the energy supplied by the time. 22. (a) Strategy Use the definition of resistance. Solution Compute the resistance (i.e., Ohm’s Law) (b) Strategy and Solution The current flows from right to left through the battery (from low to high potential). Thus, the current flows from left to right through the resistor. 23. Strategy Use Eq. (18-8). The cross-sectional areas of the wires are given by Solution Form a proportion to find the ratio of diameters. 40. (a) Strategy Use Eqs. (18-13) and (18-17). You ought to redraw the circuit so that you can see what things are in parallel or series, one pair at a time. Solution Compute the resistance between points A and B. (b) Strategy Label the currents on a diagram. Use Kirchhoff’s rules. I Solution The current through the emf is where the currents labeled 1 I1 and 2 are shown in the diagram. Applying the loop rule, we have Solve for the current through the resistor, I2 R2 = 24 Ω R1 = 12 Ω I 42. Strategy Use Eqs. (18-13) and (18-17). The current through the through the emf, resistor is the same as that Solution Find Find R. 44. (a) Strategy Use Eqs. (18-15) and (18-18). Capacitors “add up” in the opposite way that resistors do, so that when they’re in series they increase the total capacitance, and when they’re in parallel they decrease the total capacitance. Solution Find the unknown capacitance. (b) Strategy The charge on the Use Eq. (17-14). capacitor is the same charge as on the equivalent capacitor. Solution Find the charge. 52. Strategy Use Kirchhoff’s rules. Let be the top branch, be the middle branch, and be the bottom branch. Assume that each current flows right to left. NOTE: This strategy and solution is just one way of setting up the problem. You could label the currents differently and even have them go different directions, but the equations you create from these will all solve to the same result. The algebra is messy, but not technically difficult. Solution Find the current in each branch of the circuit. (1) (2) Substitute (1) into (2). Solve (3) for Solve for Calculate Calculate and substitute into (4). (3) To two significant figures, the currents are: Branch I (mA) Direction Top 85 right to left Middle 56 left to right Bottom 29 left to right 55. Strategy Use Kirchhoff’s rules. Let the current on the left be I, the one in the middle be one on the right be and the flows downward. NOTE: Same as above; the details of what you call each current and which directions you use are up to you. You just need to make sure that you’re consistent. Also note that in the printing of some of these equations, some symbols don’t always come through. “delta V” sometimes looks like a % sign. Solution Find the unknown emf and the unknown resistor. Loop ABCFA: Loop ABCDEFA: 57. Strategy Use Eq. (18-20). Solution Compute the power (P=IV) dissipated by the resistor. 60. Strategy Use Eq. (18-21b). Solution Compute the resistance of the lightbulb. 61. Strategy and Solution Yes; the power rating can be determined by