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Unformatted text preview: PHYS 2B Quiz 4 Solutions Aris February 7, 2010 1 Problem 1 From Ohm’s law J 1 = E 1 /ρ 1 . The current is I = J 1 πb 2 , i.e. the current density times the crosssectional area. The same amount of current flows through the narrow and thick cylinders. Hence I = J 1 πb 2 = J 2 π ( b/ 2) 2 . This gives J 2 = 4 * J 1 = 4 E 1 /ρ 1 . We use Ohm’s law again in the second cylinder. E 2 = ρ 2 J 2 = 4 ρ 2 E 1 /ρ 1 = 4 * 2 . 65 * 4 / 1 . 68 = 25 V/cm . 2 Problem 2 From problem 1, we know that J 2 = 4 J 1 or n 2 ev 2 = 4 n 1 ev 1 . Hence n 1 /n 2 = v 2 / (4 v 1 ) = 1 . 5 / 2 = 0 . 75. 3 Problem 3 When a potential difference of 1 . 5 V is applied across a bulb with resistance 6Ω, a current I = V/R = 1 . 5 / 6 = 0 . 25 A flows. This causes a power dissipation of I 2 R = 0 . 25 * . 25 * 6 = 0 . 375 W in the bulb. In other words, 0 . 375 J of energy is continually being drained from the battery per second. The battery lasts for 4500 / . 375 = 12000 s = 200 min = 3 . 3 h ....
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 Winter '10
 Hirsch
 Resistor, Potential difference, Electric charge, R R R

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