midterm3

# Question 14 chap 31 sect 6 part 1 of 1 10 points a

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Question 14, chap 31, sect 6. part 1 of 1 10 points A large electromagnet has an inductance of 50 H and a resistance of 6 Ω. It is connected to a DC power source of 350 V. Find the time for the current to reach 15 A. Correct answer: 2 . 4771. Explanation: Let : L = 50 H , R = 6 Ω , I = 15 A , and E o = 350 V , The current constant of this inductance is I f = ε 0 R = 350 V 6 Ω = 58 . 3333 A , and the time constant τ = L R = 50 H 6 Ω = 8 . 33333 s . If the current is initially zero in this LR cir- cuit, the current at time t is I = I f parenleftBig 1 e t/τ parenrightBig I I f = 1 e t/τ e t/τ = 1 I I f t τ = ln parenleftbigg 1 I I f parenrightbigg t = τ ln parenleftbigg 1 I I f parenrightbigg = (8 . 33333 s) ln parenleftbigg 1 15 A 58 . 3333 A parenrightbigg = 2 . 4771 s . Question 15, chap 31, sect 2. part 1 of 1 10 points A solenoid with circular cross section pro- duces a steadily increasing magnetic flux through its cross section. There is an octago- nally shaped circuit surrounding the solenoid as shown below. The increasing magnetic flux gives rise to a counterclockwise induced emf E . Initial Case: The circuit consists of two identical light bulbs of equal resistance, R , connected in series, leading to a loop equation E− 2 i R = 0.

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midterm 03 – LIN, KEVIN – Due: Apr 3 2008, 11:00 pm 13 B B B B X Y i i Figure 1: Primed Case: Now connect the points C and D with a wire CAD, (see the figure below). B B B B X Y D C A i 2 i 1 i 3 i 3 Figure 2: What happens after the points C and D are connected by a wire as in the second (primed) case? 1. Bulb Y goes out and bulb X gets dim- mer. 2. Bulb Y goes out and bulb X gets brighter. correct 3. Bulb Y goes out and bulb X remains at the same brightness. 4. Bulb X goes out and bulb Y gets brighter. 5. Bulb X goes out and bulb Y remains at the same brightness. 6. Bulb X goes out and bulb Y gets dim- mer. Explanation: Basic Concepts: Induced emf. Solution: Let E and R be the induced emf and resistance of the light bulbs, respectively. For the first case, since the two bulbs are in series, the equivalent resistance is simply R eq = R + R = 2 R and the current through the bulbs is i = E 2 R . Hence, for the first case, the power consumed by bulb X is P X = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R For the second (primed) case, since bulb Y is shorted, the current through bulb X is now i = E R , and the power consumed by bulb X is P X = parenleftbigg E R parenrightbigg 2 R = E 2 R . Hence the ratio is P X P X = E 2 R E 2 4 R = 4 . For the second (primed) case, bulb Y is shorted; i.e. , P Y = 0 . Hence the ratio is 0. Therefore bulb X gets brighter and bulb Y goes out. Question 16, chap 31, sect 5. part 1 of 1 10 points An inductor of 390 turns has a radius of 3 cm and a length of 22 cm. The permeability of free space is 1 . 25664 × 10 6 N / A 2 .
midterm 03 – LIN, KEVIN – Due: Apr 3 2008, 11:00 pm 14 Find the energy stored in it when the cur- rent is 0 . 4 A.

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