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OHW 16 - terry(ect328 oldhomework 16 Turner(59130 This...

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terry (ect328) – oldhomework 16 – Turner – (59130) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 2 . 6 × 10 6 Ω 1 . 9 μ F 10 V S An emf of 10 V is connected to a series RC circuit consisting of a 2 . 6 × 10 6 Ω resistor and a 1 . 9 μ F capacitor. Find the time required for the charge on the capacitor to reach 88% of its final value. Correct answer: 10 . 4741 s. Explanation: Let : R = 2 . 6 × 10 6 Ω C = 1 . 9 μ F = 1 . 9 × 10 6 F E = 10 V , and p = 0 . 88 . From Q = Q max (1 - e t/τ ), where τ = R C , we get p = Q Q max = 1 - e t/τ e t/τ = 1 - p - t τ = ln(1 - p ) t = - τ ln(1 - p ) = - R C ln(1 - p ) = - ( 2 . 6 × 10 6 Ω )( 1 . 9 × 10 6 F ) × ln(1 - 0 . 88) = 10 . 4741 s . keywords: 002 (part 1 of 4) 10.0 points A typical television dissipates 284 W when plugged into a 115 V outlet. Find the resistance of the television. Correct answer: 46 . 5669 Ω. Explanation: Let : P = 284 W and V = 115 V . Power is P = V 2 R so the television resistance is R T V = V 2 P T V = (115 V) 2 284 W = 46 . 5669 Ω . 003 (part 2 of 4) 10.0 points The 2 . 41 Ω fuse (integrated into the outlet) and wires connecting the outlet form a series circuit that works like a voltage divider. As- sume there are no other appliances plugged into the outlet. Find the voltage drop across the television. Correct answer: 109 . 341 V. Explanation: Let : R fuse = 2 . 41 Ω . When the television and a fuse are con- nected in series, the total resistance is R tot = R T V + R fuse This combination works like a voltage divider, where V R , so V T V R T V =
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