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Transcript

Electromagnetism G. L. Pollack and D. R. Stump What is the resistance of this network? Four stepped exercises. Exercise 1. Consider the unit cube, shown in Fig. 1(a). Suppose each edge is a conductor with resistance r. What is the total resistance between diagonal corners of the cube? General Strategy The surefire way to find the total resistance R between two terminals in any network is to let current I enter at one terminal and exit at the other. Use Kirchhoff’s laws (about which more below) together with any symmetries available to find the current through each of the conductors of the network. Finally, calculate the total voltage change ∆V for any path between the original terminals and obtain R from R = |∆V |/I. Kirchhoff ’s first law states that the current entering any junction equals the current leaving it. This law is a consequence of conservation of charge. Kirchhoff ’s second law states that the net emf, i.e., the algebraic sum of the voltage changes, around any closed path in the circuit is zero. This law is a restatement of the conservativeness of the electrostatic field—a consequence of the field equation ∇× E = 0 for electrostatics. Exercise 1 has sufficient symmetry so that only Kirchhoff’s first law need be applied. What is the resistance R between the diagonal corners, say, (0, 0, 0) and (1, 1, 1)? Let current I enter the network at (0, 0, 0) and leave at (1, 1, 1). From symmetry it is clear that three equal currents, each I/3, flow away from (0, 0, 0) along the x, y, z axes, repesectively. When these current reach a corner, they divide symmetrically into two equal currents. Thus, upon reaching the corner (0, 0, 1) the current along the z axis divides into two currents, each I/6, one from (0, 0, 1) to (0, 1, 1) and the other from (0, 0, 1) to (1, 0, 1). The currents along the x and y axes behave similarly. Consider next what happens at the junction (0, 1, 1): Two b equal currents bjI/6 and kI/6 flow into it so that current biI/3 must flow out from (0, 1, 1) to (1, 1, 1). The cube with all currents is shown in Fig 1(b). Finally then for the path (0, 0, 0) → (0, 0, 1) → (0, 1, 1) → (1, 1, 1) we have the voltage drop I I I ∆V = − r − r − r = IR 3 6 3 so that R = 5r/6. Exercise 2. Consider now the planar array of 12 resistors, each with resistance r, shown in Fig. 2(a). What is the total resistance R between the corners A, at (−1, 1), and A0 , at (1, −1)? In solving this exercise, Kirchhoff’s first law and symmetry will take us only part way. We will also need to apply the second law. Let current I enter the network at A and leave at A0 . From symmetry it is clear that two equal currents I/2 flow along AB and AD. But it is not clear, i.e., there is not enough symmetry, to determine how the current divides at junction B. Therefore call I1 the current that flows along BE and, because Kirchhoff’s first law must be satisfied, we must have I/2 − I1 flowing along BC and CF . Similarly the current along AD divides into currents I1 along DE, and I/2 − I1 along DC 0 . Figure 2(b) shows the array with labeled currents in each resistance of the network. Note the symmetry at junction E—two equal currents I1 flow in and two currents I1 flow away. Now to determine the unknown current I1 , apply Kirchhoff’s second law to a counterclockwise loop around F CBEF . Counting the voltage changes against the current direction as positive and those in the current direction as negative, we have I I − I1 r + − I1 r − I1 r − I1 r = 0, ∆V = 2 2 from which I1 = I/4. Finally then for the path A → B → C → F drop is I I I I I ∆V = − r − − r− − r− 2 2 4 2 4 so that R = 3r/2. → A0 , the voltage I r = IR, 2 The next two exercises refer to the 24 resistor array shown in Figure 3. Each segment has resistance r. Exercise 3. What is the resistance between terminals F and F 0 ? (Hint: First convince yourself that no current flows past the corners A or A0 . However, your solution should also show that I/14 flows past the corners D and D 0 .) [Answer: 5r/7] Exercise 4. What is the resistance between the terminals A and A0 ? [Answer: 13r/7] This is a hard problem.