Phy-131 Practice test solution - Test 2 PDF

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Download Phy-131 Practice test solution - Test 2 PDF

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Phy-131 Practice test solution: Test 2 1. Draw a dx at location x between x=0 and x=3 m. The distance from dx to point P at 4 m is 4 m - x. The integral is of kdq/dist for all the dq's from x=0 to x=3 m. Replace dq with lambda*dx where lambda is 3 nC/m. k*lambda=27 N*m/C = 27 V. 2. In electrostatic equilibrium, the rule is "One conductor, one potential." 3. 1, 2, 5, and 6 all have the battery EMF across a single resistor, and so tie. 3 and 4 have the EMF across two of the resistors in parallel (which halves the resistance); so currents 3 and 4 are double the others. 4. A has EMF over 1 bulb. In B the bulb is shorted out, and has no current. In C, the EMF is over 2 bulbs, so less current than in A. 5. A has EMF over 1 bulb. B has EMF over 2 in parallel, thus twice the current as in A. C has EMF of 2 bulbs, thus half the current in A. 6. A is battery EMF. B is half battery EMF. C is 1/3 battery EMF. 7. In A, the Reff is 3 bulbs. In B, the Reff is 1.5 bulbs. In C, the Reff is 1/3 bulb. 8. Use series or parallel analysis for R's. 6 and 12 in parallel=4 Ohms, so the upper branch in 8 Ohms, as is the lower, so the total Reff is 4 Ohms. The current through the EMF is 3 A and 1.5 A goes in each branch; so 1.5 A through 5 Ohm R. 9. Use series or parallel analysis for R's. 20, 10, and 10 in parallel yield 4 Ohms, so the total is 6 Ohms. The current through the EMF is 2 A, so the voltage across Reff is 8 V. So current through 10 Ohm is 0.8 A. 10. Find the current, then use units to get the number of e's. Current is 1.5 C/s. There are 1.6E-19 C/e, so 9.38E18 e/s, so 5.6E21 e in 600 s. 11. CAP is PROP to area, and INV PROP to separation distance. Ignore the circuit info for this part. Doubling DIST halves CAP. Now use the circuit info. The VOLTAGE was constant while the CAP was halved, so

half as much charge is separated (the other half "went back home". 12. Use series or parallel analysis for C's. 10 and 30 in parallel yield 40 muF, and 40 and 20 in series yield 40/3 muF. 18 V applied to that Ceff separates 240 muC. Since the POS plate of C1 is the POS plate of Ceff, thus C1 separates 240 muC=.24 mC. 13. The voltage across A is half the EMF; the same is true for B, C, D. The voltage across E is all the EMF, so E is brightest. 14. For clarity, redraw the circuit so all the current will pass through a single wire in the middle of the 6 R's. Then the first 3 R's consist of an upper branch with 1 R and a lower branch with 2 R's in series, for an Reff of 2/3 R. The second 3 R's have the upper branch with 2 R's in series and a lower branch with a single R, so also Reff of 2/3 R. So the total Reff for all 6 is 4/3 R. Connect to a 48 V EMF and imagine current left to right, then 3 A total splits into 2 A (upper) and 1 A (lower); at the central connection 1 A flows towards bottom of page; then 2 A (lower) and 1 A (upper) finish the journey to b. 15. Use series or parallel analysis for R's. The 3 in series yield 30 Ohms. 30 and 20 in parallel yield 12 Ohms, so 17 Ohms is Reff. So CURR through EMF is 11/17 A, so voltage across 5.0 Ohms is 55/17 V, and voltage across 20 Ohms is 132/17 V=7.76 V. 16. Simplest version is to square the RES voltage and divide by the resistance, before the 120 s multiplication. Or, find current, then use I*V or I^2R, before the multiplication. All methods yield 32 J. 17. Use series or parallel analysis for C's. The 2 in parallel yield 2C, then the 3 in series yield 0.5 C for Ceff. 18. The first sentence gives C. For the energy stored use any of the three equivalent expression 0.5QV or 0.5Q^2/C or 0.5CV^2. The last is the simplest to use here. All yield 4.41 mJ. 19. Use series or parallel analysis for R's. 2 R's in series yield 2R which is in parallel with 2R which yields R for Reff. Now R and R are in series for the total Reff=2R= 6 Ohms. So the current through the EMF is 2 A, which splits equally in the parallel branches. So

1 A is passing through the 2R resistor for I^2R of 6.0 W. 20. Use LHR since charge is negative. If thumb is +x, and palm is +z, then fingers must be -y. 21. Use Lorentz mag force. vcrossB must be equal to 2i-10j+6 in units of N/C and Bx=0, so (2 m/s)By = 6 N/C so By = 3 T, -(2 m/s)Bz = -10 M/C so Bz = 5 T, and (4 m/s)Bz-(6 m/s)By = 2 N/C is a check. 22. Use LHR since charge is negative. In panel 1, fingers are out of page. In 2, fingers are left. In 3, fingers are towards top of page. In 4, B is either into or out of page. So only panel 1, has B definitely out of page. 23. Use Lorentz mag force, and find qvBsin(theta), where q, v, and B, are all the same. For 3, theta=0, so sin=0. For 1, theta=90 so sin is maximum at 1. For 2, sin is between 0 and 1....