L l d i dt at all times so v l 0 as t and we can

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L = L d I dt , at all times, so V L 0 as t → ∞ , and we can replace the inductor in the figure by a straight wire. When the current now comes from R 1 to the junction where I 1 splits into I 2 and I , there is no resistance in the “ I ” path but a nonzero resistance R 2 in the other. Naturally, the current takes the path with no resistance. Since we do pass through R 1 in any case, the equivalent resistance is now R 1 . At t = I = E R 1 and I 2 = 0 Alternate Solution: An equation of the form L dI dt R I − E = 0 has a solution which is composed of a ho- mogeneous solution and a particular solution. The particular solution corresponds to the first term in the I in the problem statement; the homogeneous corresponds to the second term. The homogeneous solution is therefore the only one of relevance to the time con- stant τ x , and we can disregard any batteries
Faraday’s Law and AC circuits yeazell (58010) 8 E . This provides a simple solution to Part 3: Just ignore the battery and find the equiva- lent resistance connected to L , in this case 020(part2of2)10.0points
4. I 2 = E R 2 Explanation: See previous explanation. 019(part1of2)10.0points An LC circuit is shown in the figure below. The 27 pF capacitor is initially charged by the 6 V battery when S is at position a . Then S is thrown to position b so that the capacitor is shorted across the 12 mH inductor. 12 mH 27 pF 6 V S b a What is the maximum value for the oscil- lating current assuming no resistance in the circuit? What is the maximum energy stored in the magnetic field of the inductor? Correct answer: 4 . 86 × 10 - 10 J. Explanation: U max,ind = U max,cap = Q 2 max 2 C = 1 2 C V 2 = 1 2 (2 . 7 × 10 - 11 F) (6 V) 2 = 4 . 86 × 10 - 10 J . 021 10.0points The emf E drives the circuit shown below at angular frequency ω. E C R L
Faraday’s Law and AC circuits yeazell (58010)

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