Midterm_2-solutions.pdf

# Midterm_2-solutions.pdf - Version 028 Midterm 2...

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Version 028 – Midterm 2 – turner – (90130) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A 2 . 5 μ F spherical capacitor is composed of two metal spheres, one having a radius twice as large as the other. If the region between the spheres is a vac- uum, determine the volume of this region. 1. 3541580000000.0 2. 10898800000000.0 3. 24642100000000.0 4. 41575700000000.0 5. 2660840000000.0 6. 87190500000000.0 7. 28332600000000.0 8. 58410800000000.0 9. 32374500000000.0 10. 7301350000000.0 Correct answer: 4 . 15757 × 10 13 m 3 . Explanation: Let : C = 2 . 5 μ F = 2 . 5 × 10 6 F and ǫ 0 = 8 . 8542 × 10 12 C 2 / N · m 2 . Let the radii be a and 2 a . Put a charge Q on the inner conductor and Q on the outer conductor. The potential of the inner conduc- tor is V inner = k e Q a while that of the outer is V outer = k e Q 2 a , so the potential difference is Δ V = V inner V outer = k e Q 2 a C = Q Δ V = 8 π ǫ 0 a a = C 8 π ǫ 0 = 2 . 5 × 10 6 F 8 π (8 . 8542 × 10 12 C 2 / N · m 2 ) = 11234 . 4 m , and the volume between the conductors is ΔVol = 4 3 π (2 a ) 3 4 3 π ( a ) 3 = 4 3 π ( 7 a 3 ) = 4 3 π bracketleftbig 7 (11234 . 4 m) 3 bracketrightbig = 4 . 15757 × 10 13 m 3 . 002 10.0points A 16 V battery delivers 107 mA of current when connected to a 73 Ω resistor. Determine the internal resistance of the battery. 1. 109.275 2. 19.0171 3. 76.5327 4. 97.0672 5. 23.0 6. 62.6154 7. 83.7143 8. 45.6471 9. 87.9048 10. 24.6949 Correct answer: 76 . 5327 Ω. Explanation: Let : V = 16 V , I = 107 mA = 0 . 107 A , and R = 73 Ω . The internal resistance is in series with the given resistor, so V = I ( R + r ) r = V I R = 16 V 0 . 107 A 73 Ω = 76 . 5327 Ω . 003 10.0points All the bulbs in the figure below have the same resistance R . The switch S is initially closed.

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Version 028 – Midterm 2 – turner – (90130) 2 V i A i C i B i D S i 0 If bulb B is removed from the circuit, i.e. , the switch S is opened, what happens to the currents through 1) the battery, 2) bulb A , and 3) bulb D ; Notice that in the diagram the current through the battery, i battery , is labeled as i 0 . Hint: You may find it helpful to work out the currents through bulb A , bulb D , and the battery for both cases by using V = 1 volt and R = 1 Ω. 1. i A decreases, i D remains the same, i battery decreases correct 2. i A remains the same, i D remains the same, i battery remains the same 3. i A decreases, i D decreases, i battery de- creases 4. i A increases, i D increases, i battery remains the same 5. i A decreases, i D decreases, i battery remains the same 6. i A increases, i D increases, i battery de- creases 7. i A increases, i D decreases, i battery de- creases 8. i A remains the same, i D increases, i battery increases 9. i A increases, i D increases, i battery in- creases 10. i A increases, i D remains the same, i battery increases Explanation: 1 R BC = 1 R + 1 R = 2 R R BC = R 2 R L = R A + R BC = R + R 2 = 3 R 2 1 R eq = 1 R L + 1 R = 2 3 R + 1 R = 5 3 R R eq = 3 R 5 Therefore i A = V 3 R 2 = 2 V 3 R i D = V R i battery = 5 3 V R WithoutbulbB: R L = 2 R 1 R eq = 1 2 R + 1 R = 3 2 R R eq = 2 R 3 Therefore i A = V 2 R i D = V R i battery = 3 2 V R Qualitatively, if bulb
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