HW04_Solutions - HOMEWORK 4: SOLUTIONS Q25.2: ANSWER: there...

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25-1 HOMEWORK 4: SOLUTIONS Q25.2: ANSWER: there is no electric field in a static case. When electrons are moving, there is an electric field. Q25.23: ANSWER: in case of short circuit, the current is given by the voltage over internal resistance. Therefore, larger internal resistance minimizes current and therefore dissipated power. 25.30. IDENTIFY: When the ohmmeter is connected between the opposite faces, the current flows along its length, but when the meter is connected between the inner and outer surfaces, the current flows radially outward. (a) SET UP: For a hollow cylinder, R = L/A , where A = π ( b 2 a 2 ). EXECUTE:     8 22 2.75 10 m (2.50 m) / (0.0460 m) (0.0320 m) L R L A ba    = 2.00 5 10 Ω (b) SET UP: For radial current flow from r = a to r = b , R = ( /2 πL ) ln( b/a ) (Example 25.4) EXECUTE: 8 m 4.60 cm ln( / ) ln 2 2 (2.50 m) 3.20 cm R b a L      = 6.35 10 10 Ω EVALUATE: The resistance is much smaller for the radial flow because the current flows through a much smaller distance and the area through which it flows is much larger. 25.34. (a) IDENTIFY: The idealized ammeter has no resistance so there is no potential drop across it. Therefore it acts like a short circuit across the terminals of the battery and removes the 4.00-Ω resistor from the circuit. Thus the only resistance in the circuit is the 2.00-Ω internal resistance of the battery. SET UP: Use Ohm’s law: I = E /r. EXECUTE: I = (10.0 V)/(2.00 Ω) = 5.00 A. (b) The zero-resistance ammeter is in parallel with the 4.00-Ω resistor, so all the current goes through the ammeter. If no current goes through the 4.00-Ω resistor, the potential drop across it must be zero.
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HW04_Solutions - HOMEWORK 4: SOLUTIONS Q25.2: ANSWER: there...

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