tsl336 - 2 Solution: (a) I C = 0 , I 2 = E R 1 + R 2 = 12V...

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Intermediate Exam II: Problem #1 (Spring ’05) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance C eq . (b) Find the voltage V 3 across capacitor C 3 . (c) Find the the charge Q 2 on capacitor C 2 . 8V F μ = 2 2 C F μ C 1 F μ = 3 3 C = 1 Solution: (a) C 12 = C 1 + C 2 = 3 μ F , C eq = 1 C 12 + 1 C 3 « - 1 = 1 . 5 μ F . (b) Q 3 = Q 12 = Q eq = C eq (8V) = 12 μ C V 3 = Q 3 C 3 = 12 μ C 3 μ F = 4V . (c) Q 2 = V 2 C 2 = 8 μ C . tsl336 – p.1/4
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Intermediate Exam II: Problem #2 (Spring ’05) Consider the electrical circuit shown. (a) Find the equivalent resistance R eq . (b) Find the current I 3 through resistor R 3 . 12V 6Ω R = 3 3Ω R = 2 2Ω R = 6 Solution: (a) R 36 = 1 R 3 + 1 R 6 « - 1 = 2Ω , R eq = R 2 + R 36 = 4Ω . (b) I 2 = I 36 = 12V R eq = 3A V 3 = V 36 = I 36 R 36 = 6V I 3 = V 3 R 3 = 2A . tsl337 – p.2/4
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Intermediate Exam II: Problem #3 (Spring ’05) This RC circuit has been running for a long time. (a) Find the current I 2 through the resistor R 2 . (b) Find the voltage V C across the capacitor. C = 7nF 12V R = 2Ω 1 4Ω R =
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Unformatted text preview: 2 Solution: (a) I C = 0 , I 2 = E R 1 + R 2 = 12V 6Ω = 2A . (b) V C = V 2 = I 2 R 2 = (2A)(4Ω) = 8V . tsl338 – p.3/4 Intermediate Exam II: Problem #4 (Spring ’05) Consider a charged particle moving in a uniform magnetic field as shown. The velocity is in y-direction and the magnetic field in the yz-plane at 30 ◦ from the y-direction. (a) Find the direction of the magnetic force acting on the particle. (b) Find the magnitude of the magnetic force acting on the particle. + 30 o B = 4mT v = 3m/s y z q = 5nC Solution: (a) Use the right-hand rule: positive x-direction (front, out of page). (b) F = qvB sin 30 ◦ = (5 × 10-9 C)(3m / s)(4 × 10-3 T)(0 . 5) = 3 × 10-11 N . tsl339 – p.4/4...
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This note was uploaded on 04/30/2008 for the course PHYS 204 taught by Professor Andrevantonder during the Spring '07 term at Rhode Island.

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tsl336 - 2 Solution: (a) I C = 0 , I 2 = E R 1 + R 2 = 12V...

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