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Unformatted text preview: Physics 213 HW #4 Solutions Spring 2008 25.28. IDENTIFY: R T = R [1 + ( T T )] SET UP: R = 217.3 . R T = 215.8 . For carbon, = 0.00050 (C ) 1 . EXECUTE: T T = ( R T / R ) 1 = (215.8 / 217.3 ) 1 0.00050 (C ) 1 = 13.8 C . T = 13.8 C + 4.0 C = 17.8 C. EVALUATE: For carbon, is negative so R decreases as T increases. 25.37. IDENTIFY: The voltmeter reads the potential difference V ab between the terminals of the battery. SET UP: open circuit I = 0. The circuit is sketched in Figure 25.37a. ( stands for battery emf.) EXECUTE: V ab = = 3.08 V Figure 25.37a SET UP: switch closed The circuit is sketched in Figure 35.37b. EXECUTE: V ab = Ir = 2.97 V r = 2.97 V I r = 3.08 V 2.97 V 1.65 A = 0.067 Figure 25.37b And V ab = IR so R = V ab I = 2.97 V 1.65 A = 1.80 . EVALUATE: When current flows through the battery there is a voltage drop across its internal resistance and its terminal voltage V is less than its emf. 25.45. IDENTIFY: A 100-W European bulb dissipates 100 W when used across 220 V. (a) SET UP: Take the ratio of the power in the US to the power in Europe, using P = V 2 / R . EXECUTE: P US P E = V US 2 / R V E 2 / R = V US V E 2 = 120 V 220 V 2 . This gives P US = (100 W) 120 V 220 V 2 = 29.8 W. (b) SET UP: Use P = IV to find the current. EXECUTE: I = P/V = (29.8 W)/(120 V) = 0.248 A EVALUATE: The bulb draws considerably less power in the U.S., so it would be much dimmer than in Europe. 25.86. IDENTIFY: The power output of the source is VI = ( Ir ) I . SET UP: The short-circuit current is I short circuit = / r ....
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