HW04-solutions - gilbert (amg3448) HW04 tsoi (57210) This...

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gilbert (amg3448) – HW04 – tsoi – (57210) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points AFter a 3 . 6Ωre s i s to ri sconnec tedac ro s sa battery with a 0 . 13 Ω internal resistance, the electric potential between the physical bat- tery terminals is 6 V. What is the rated emF oF the battery? Correct answer: 6 . 21667 V. Explanation: Let : R =3 . , r =0 . 13 Ω , and V =6V . The current drawn by the external resistor is given by I = V R = 6V 3 . =1 . 66667 A . The output voltage is reduced by the inter- nal resistance oF the battery by V = E- Ir, so the electromotive Force is E = V + Ir =6V+(1 . 66667 A) (0 . 13 Ω) = 6 . 21667 V . 002 10.0 points Consider the circuit shown in the fgure. In this circuit, r i is the internal resistance oF the battery (it cannot be removed and must be considered as part oF the battery); R o is an external resistance; a and b are the battery terminals; S is a switch. E 0 r i R o S a b c ±ind the power dissipated at R o when the switch is closed. 1. P = E 2 o r i 2. P = ± E o r i ² 2 R o 3. P = E 2 o R o + r i 4. P = ± E o R o ² 2 r i 5. P = E 2 o R o 6. P = E 2 o R o 7. P = ± E o R o + r i ² 2 R o correct 8. P = E 2 o r i Explanation: The power dissipated at R o is given by P = I 2 R o = ± E o R o + r i ² 2 R o . 003 10.0 points Al eng tho Fw i r ei scu tin to6equa lp i e c e s . The 6 pieces are then connected parallel, with the resulting resistance being 4 Ω. What was the resistance r oF the original length oF wire? Correct answer: 144 Ω. Explanation: Let : n =6 and R p =4Ω . The resulting resistance R p oF n equal piece
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gilbert (amg3448) – HW04 – tsoi – (57210) 2 resistors of resistance r connected parallel is R p = r n , but when they are in series, the total resis- tance R s is R s = nr = n 2 R p =6 2 (4 Ω) = 144 Ω . 004 (part 1 of 8) 10.0 points Unlike most real bulbs, the resistances of the bulbs in the questions below do not change as the current through them changes. All bulbs considered in this problem are identical. Assume all batteries are ideal (they have no internal resistance) and connecting wires have no resistance. What is true when one bulb is brighter than another? 1. Not enough information is given. 2. The current passing through both bulbs are the same. 3. The current passing through the brighter bulb is larger. correct 4. The current passing through the brighter bulb is smaller. Explanation: The brightness depends on the power P = I 2 R. When one bulb is brighter, it has a larger P . Since the bulbs are identical, the resistances are equal, so the current I must be larger in the brighter bulb. 005 (part 2 of 8) 10.0 points Al igh tbu lbandabatteryareconnectedas shown in Figure 1. E A B C D Figure 1 What is true about the current passing through various points in this circuit? 1. The current passing through point B is largest.
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This note was uploaded on 03/22/2012 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.

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HW04-solutions - gilbert (amg3448) HW04 tsoi (57210) This...

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