This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: yang (ey942) – ohw21 – turner – (56705) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A circuit is set up as shown in the figure. L I 2 R 2 I 1 R 1 E S The switch S in the circuit shown has been closed for a long time. At t = 0, S is opened. What is the voltage V 1 across resistor R 1 as a function of time t ? 1. V 1 = parenleftbigg R 1 R 1 + R 2 E parenrightbigg e − R 1 R 2 t/ [( R 1 + R 2 ) L ] 2. V 1 = E e − R 1 R 2 t/ [( R 1 + R 2 ) L ] 3. V 1 = E e − R 2 t/L 4. V 1 = E e − ( R 1 + R 2 ) t/L 5. V 1 = parenleftbigg R 1 R 2 E parenrightbigg e − R 1 R 2 t/ [( R 1 + R 2 ) L ] 6. V 1 = parenleftbigg R 1 R 2 E parenrightbigg e − R 2 t/L 7. V 1 = parenleftbigg R 1 R 1 + R 2 E parenrightbigg e − ( R 1 + R 2 ) t/L 8. V 1 = parenleftbigg R 1 R 2 E parenrightbigg e − ( R 1 + R 2 ) t/L correct Explanation: Just before the switch is opened ( i.e. , t ≤ 0) the current through the inductor L and resistor R 2 is I max = E R 2 , (1) since the inductor has an inductance of zero at this moment and the potential difference across the L and R 2 is the emf supplied by the battery E . When the switch is opened ( t = 0) the current in the top loop remains I max for an instant. The electrical potential drop V max 1 across resistor R 1 is V max 1 = I max R 1 , so (2) V max 1 = R 1 R 2 E V 1 = V max 1 e − t/τ τ = L R total = L R 1 + R 2 V 1 = parenleftbigg R 1 R 2 E parenrightbigg e − ( R 1 + R 2 ) t/L . 002 (part 1 of 2) 10.0 points A long insulated wire with a resistance of 21 Ω / m is to be used to construct a resistor. First, the wire is bent in half and then the doubled wire is wound in a cylinderical form, as show in figure. The diameter of the cylin drical form is 4 cm, its length is 20 cm, and the total length of wire is 15 m. Find the inductance of this wirewound reisitor. Correct answer: 0 H. Explanation: Let : ℓ = 20 cm = 0 . 2 m , d = 4 cm = 0 . 04 m , r = 21 Ω / m , and L = 15 m . The total flux in the coil is zero, so L = 0 H . 003 (part 2 of 2) 10.0 points Find the resistance of this wirewound resis tor. yang (ey942) – ohw21 – turner – (56705) 2 Correct answer: 315 Ω. Explanation: Let : r = 21 Ω / m and L = 15 m . The total resistance of the wire is R = r L = (21 Ω / m) (15 m) = 315 Ω . keywords: 004 (part 1 of 2) 10.0 points A solenoid has 144 turns of wire uniformly wrapped around an airfilled core, which has a diameter of 14 mm and a length of 10 . 5 cm. The permeability of free space is 1 . 25664 × 10 − 6 N / A 2 . Calculate the selfinductance of the solenoid. Correct answer: 3 . 82025 × 10 − 5 H....
View
Full Document
 Spring '10
 Turner
 Energy, Magnetic Field, Correct Answer, electrical energy

Click to edit the document details