The figure below shows a resistor of resistance R = 6.29 ? connected to an ideal
ID: 2209400 • Letter: T
Question
The figure below shows a resistor of resistance R = 6.29 ? connected to an ideal battery of emf = 15.1 V by means of two copper wires. Each wire has length 20.7 cm and radius 1.50 mm. In dealing with such circuits in this chapter, we generally neglect the potential differences along the wires and the transfer of energy to thermal energy in them. Check the validity of this neglect for the circuit of the figure below. What is the potential difference across (a) the resistor and (b) each of the two sections of wire? At what rate is energy lost to thermal energy in (c) the resistor and (d) each section of wire?
The figure below shows a resistor of resistance R = 6.29 ? connected to an ideal battery of emf = 15.1 V by means of two copper wires. Each wire has length 20.7 cm and radius 1.50 mm. In dealing with such circuits in this chapter, we generally neglect the potential differences along the wires and the transfer of energy to thermal energy in them. Check the validity of this neglect for the circuit of the figure below. What is the potential difference across (a) the resistor and (b) each of the two sections of wire? At what rate is energy lost to thermal energy in (c) the resistor and (d) each section of wire?Explanation / Answer
resistance of copper = 17.2*10^-9*.207/(3.14*.0015^2) = 5.04*10^-4 = 0.0005ohm so, net resistance = 6.29+2*0.0005 = 6.291 ohm so, net current = 15.1/6.291 = 2.4 A potential difference across a) resistor = 2.4*6.29 = 15.098 V b) each of the two sections = 1.2 mV c) energy lost in the resistor = I^2*R = 2.4^2*6.29 = 36.24J d) energy lost in each section = 2.9 mW
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