RCW3%20solutions

RCW3%20solutions - Physics 204, morning summer 2008...

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Unformatted text preview: Physics 204, morning summer 2008 Recitation ClassWork 4 solutions Chapter 20 Conceptual Questions #5,6 Problems #3,28 Concepts & Calculations #118,121 5. REASONING AND SOLUTION One electrical appliance operates with a voltage of 120 V, while another operates with a voltage of 240 V. The power used by either appliance is given by Equation 20.6c: P = V 2 / R . Without knowing the resistance R of each appliance, no conclusion can be reached as to which appliance, if either, uses more power. 6. REASONING AND SOLUTION Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. The power used by either bulb is given by Equation 20.6c: P = V 2 / R . We see from Equation 20.6c that, at constant voltage, the power used by a bulb is inversely proportional to the resistance of the filament. Therefore, the filament resistance is greater for the 75‐W bulb. 3. REASONING AND SOLUTION First determine the total charge delivered to the battery using Equation 20.1: ⎛ 3600 s ⎞ 5 Δq = I Δt = (6.0 A)(5.0 h) ⎜ ⎟ = 1.1× 10 C ⎝ 1h ⎠ To find the energy delivered to the battery, multiply this charge by the energy per unit charge (i.e., the voltage) to get Energy = (Δq)V = (1.1 × 105 C)(12 V) = 1.3 × 106 J 28. REASONING A certain amount of time t is needed for the heater to deliver the heat Q required to raise the temperature of the water, and this time depends on the power produced by the heater. The power P is the energy (heat in this case) per unit time, so the time is the heat divided by the power or t = Q/P. The heat required to raise the temperature of a mass m of water by an amount ΔT is given by Equation 12.4 as Q = cmΔT, where c is the specific heat capacity of water [4186 J/(kg∙Cº), see Table 12.2]. The power dissipated in a resistance R is given by Equation 20.6c as P = V 2 / R , where V is the voltage across the resistor. Using these expressions for Q and P will allow us to determine the time t. SOLUTION Substituting Equations 12.4 and 20.6c into the expression for the time and recognizing that the normal boiling point of water is 100.0 ºC, we find that t= Q cmΔT RcmΔT = = P V2 /R V2 = (15 Ω ) ⎡ 4186 J/ ( kg ⋅ C° )⎤ ( 0.50 kg )(100.0 °C − 13 °C ) ⎣ ⎦ = 190 s (120 V )2 118. CONCEPT QUESTIONS 2 a. The power delivered to a resistor is given by Equation 20.6c as P = V / R , where V is the voltage and R is the 2 resistance. Because of the dependence of the power on V , doubling the voltage has a greater effect in increasing the power than halving the resistance. The table shows the power for each circuit, given in terms of these variables: Power Rank a b c d V2 R V2 P= 2R 2 ( 2V ) 4V 2 P= = R R 2 ( 2V ) 2V 2 P= = 2R R P= 3 4 1 2 b. The current is given by Equation 20.2 as I = V/R. Note that the current, unlike the power, depends linearly on the voltage. Therefore, either doubling the voltage or halving the resistance has the same effect on the current. The table shows the current for the four circuits: Current Rank V R V I= 2R 2V I= R 2V V I= = 2R R I= a b c d 2 3 1 2 SOLUTION a. Using the results from part (a) and the values of V = 12.0 V and R = 6.00 , the power dissipated in each resistor is Power Rank a V 2 (12.0 V ) P= = = 24.0 W 6.00 Ω R 3 b 2 (12.0 V ) V P= = = 12.0 W 2 R 2 ( 6.00 Ω ) 4 c 4V 2 4 (12.0 V ) P= = = 96.0 W R ( 6.00 Ω ) 1 d 2 2 (12.0 V ) 2V P= = = 48.0 W R ( 6.00 Ω ) 2 2 2 2 2 b. Using the results from part (b) and the values of V = 12.0 V and R = 6.00 , the current in each circuit is Current Rank a b c d V 12.0 V = = 2.00 A R 6.00 Ω 12.0 V V I= = = 1.00 A 2 R 2 ( 6.00 Ω ) I= 2V 2 (12.0 V ) = = 4.00 A R 6.00 Ω 2V 2 (12.0 V ) I= = = 2.00 A 2 R 2 ( 6.00 Ω ) I= 2 3 1 2 121. CONCEPT QUESTION In part a, the current goes from left‐to‐right through the resistor. Since the current always goes from a higher to a lower potential, the left end of the resistor is + and the right end is –. In part b, the current goes from right‐to‐left through the resistor. The right end of the resistor is + and the left end is –. The potential drops and rises for the two cases are: Potential drops Potential rises Part a IR V Part b V IR SOLUTION Since the current I goes from left‐to‐right through the 3.0‐ and 4.0‐ resistors, the left end of each resistor is + and the right end is –. The current goes through the 5.0‐ resistor from right‐to‐left, so the right end is + and the left end is –. Starting at the upper left corner of the circuit, and proceeding clockwise around it, Kirchhoff’s loop rule is written as ( 3.0 Ω ) I + 12 V + ( 4.0 Ω ) I + ( 5.0 Ω ) I = 36 V Potential drops Potential rises Solving this equation for the current gives I = 2.0 A ...
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