# HW 16 - yang(ey942 – ohw16 – turner –(56705 1 This...

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Unformatted text preview: yang (ey942) – ohw16 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The switch S has been in position b for a long period of time. C R 1 R 2 R 3 E S b a When the switch is moved to position “a”, find the characteristic time constant. 1. τ = 1 √ R 1 R 2 C 2. τ = ( R 1 + R 2 ) C correct 3. τ = 2 ( R 1 + R 2 ) C 4. τ = R 1 C 5. τ = 1 R 2 C 6. τ = 1 R 1 C 7. τ = R 1 + R 2 2 C 8. τ = R 2 C 9. τ = radicalbig R 1 R 2 C 10. τ = 1 ( R 1 + R 2 ) C Explanation: In charging an R C circuit, the characteris- tic time constant is given by τ = R C , where in this problem R is the equivalent resistance, or R = R 1 + R 2 . 002 (part 2 of 2) 10.0 points 1 . 3 μ F 1 MΩ 4 MΩ 3 . 9 MΩ 1 . 2 V S b a S has been left at position “ a ” for a long time. It is then switched from “ a ” to “ b ” at t = 0. Determine the energy dissipated through the resistor R 2 alone from t = 0 to t = ∞ . Correct answer: 0 . 473924 μ J. Explanation: Let : E = 1 . 2 V , R 1 = 1 MΩ , R 2 = 4 MΩ , R 3 = 3 . 9 MΩ , and C = 1 . 3 μ F . The total energy dissipated U R 2 + R 3 dissip = 1 2 C E 2 . Since R 2 and R 3 are in series, the energy dissipated by R 2 is only a fraction R 2 R 2 + R 3 of the total energy: U R 2 dissip = parenleftbigg R 2 R 2 + R 3 parenrightbiggparenleftbigg 1 2 C E 2 parenrightbigg We observe that power consumption consid- eration provides an independent check on the fraction used. Since the two resistors are in series they share a common current, I . The corresponding power consumptions by R 2 and R 3 are respectively P 2 = I 2 R 2 and P 3 = I 2 R 3 . This shows the correctness of the fraction, i.e. P 2 / ( P 2 + P 3 ) = R 2 / ( R 2 + R 3 ). Alternate solution: yang (ey942) – ohw16 – turner – (56705) 2 More formally, noting that the initial current is I = E R 2 + R 3 , the total energy dissipated by R 2 is U R 2 = integraldisplay ∞ I ( t ) 2 R 2 dt = integraldisplay ∞ I 2 R 2 e − 2 t/ [ C ( R 2 + R 3 )] dt = parenleftbigg E R 2 + R 3 parenrightbigg 2 R 2 bracketleftbigg − C ( R 2 + R 3 ) 2 bracketrightbigg × e − 2 t/ [ C ( R 2 + R 3 )] vextendsingle vextendsingle vextendsingle vextendsingle ∞ = R 2 R 2 + R 3 1 2 C E 2 = (4 MΩ) (4 MΩ) + (3 . 9 MΩ) 1 2 (1 . 3 μ F) (1 . 2 V) 2 = . 473924 μ J . 003 (part 1 of 2) 10.0 points Assume: The battery is ideal (it has no internal resistance) and connecting wires have no resistance. Unlike most real bulbs, the resistance of the bulb in the questions below does not change as the current through it changes. A battery, a capacitor, a bulb, and a switch are in the circuit as shown below. The switch is initially open as shown in the diagram, and the capacitor is uncharged....
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HW 16 - yang(ey942 – ohw16 – turner –(56705 1 This...

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