# exam 2 - 2Ω-resistor I = 12V 4Ω = 3A(b No current...

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Intermediate Exam II: Problem #1 (Spring ’06) The circuit of capacitors connected to a battery is at equilibrium. (a) Find the charge Q 3 on capacitor C 3 . (b) Find the charge Q 2 on capacitor C 2 . 12V C 1 = 2 μ F F μ = 2 2 C 3 C = 3 μ F Solution: (a) Q 3 = C 3 (12V) = (3 μ F)(12V) = 36 μ C . (b) Q 2 = Q 12 = C 12 (12V) = (1 μ F)(12V) = 12 μ C . tsl351 – p.1/4

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Intermediate Exam II: Problem #2 (Spring ’06) Consider the two-loop circuit shown. (a) Find the current I 1 . (b) Find the current I 2 . 2V 10V 2Ω 2Ω 2Ω 2Ω 3Ω I 2 I 1 Solution: (a) - (2Ω)( I 1 ) + 10V - (2Ω)( I 1 ) - 2V = 0 I 1 = 8V 4Ω = 2A . (b) - (2Ω)( I 2 ) + 10V - (2Ω)( I 2 ) - (3Ω)( I 2 ) = 0 I 2 = 10V 7Ω = 1 . 43A . tsl352 – p.2/4
Intermediate Exam II: Problem #3 (Spring ’06) In this RC circuit the switch S is initially open as shown. (a) Find the current I right after the switch has been closed. (b) Find the current I a very long time later. 5nF 12V I S 2Ω 4Ω Solution: (a) No current through

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Unformatted text preview: 2Ω-resistor: I = 12V 4Ω = 3A . (b) No current through capacitor: I = 12V 6Ω = 2A . tsl353 – p.3/4 Intermediate Exam II: Problem #4 (Spring ’06) A current loop in the form of a right triangle is placed in a uniform magnetic field of magnitude B = 30 mT as shown. The current in the loop is I = 0 . 4 A in the direction indicated. (a) Find magnitude and direction of the force ~ F 1 on side 1 of the triangle. (b) Find magnitude and direction of the force ~ F 2 on side 2 of the triangle. 20cm 20cm 3 1 2 B Solution: (a) ~ F 1 = I ~ L × ~ B = 0 (angle between ~ L and ~ B is 180 ◦ ). (b) F 2 = ILB = (0 . 4A)(0 . 2m)(30 × 10-3 T) = 2 . 4 × 10-3 N . Direction of ~ F 2 : ⊗ (into plane). tsl354 – p.4/4...
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exam 2 - 2Ω-resistor I = 12V 4Ω = 3A(b No current...

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