View Question
 ```Neglecting Wires Figure 27-17 shows a R = 6.20 resistor connected to an ideal 12.0 V battery by means of two copper wires. The wires each have length 19.5 cm and radius 1.00 mm. In such circuits we generally neglect the potential differences along wires and the transfer of energy to thermal energy in them. Check the validity of this neglect for the circuit of Fig. 27-17 by answering the following questions. __________ | | | | --- > - = 12V > = Resistor | > |_________| Fig. 28N-4 (a) What is the potential difference across the resistor? (Enter your answer to 6 significant figures for comparison.) _____V (b) What is it across each of the two sections of wire? ____mV (c) At what rate is energy lost to thermal energy in the resistor? _____ W (d) At what rate is it lost in each of the two sections of wire? ______ W```
 ```Hi!! To answer this question you need to find the resistance of each copper wire. The resistivity (rho) of the copper at room temperature (aprox 20°C) is 0.00000001678 ohm.m (1.678 * 10^-8). You must know that resistance R = resistivity * length / area ; the length of each wire L is 0.195 m and the cross section area A is: A = PI * r^2 = 3.141593 * 1^2 mm^2 = 3.141593 mm^2 = 0.000003141593 m^2 Now: R = 0.00000001678 ohm.m * 0.195 m / 0.000003141593 m^2 = = 0.0010415416637355634 ohm Total resistance of the circuit Rt is: Rt = 2 * 0.0010415416637355634 ohm + 6.20 ohm = = 6.202083 ohm Then the current I on the circuit is: I = V / Rt = 12 V / 6.202083 ohm = 1.934834 A Now we can answer the questions: (a) What is the potential difference across the resistor? (Enter your answer to 6 significant figures for comparison.) V(resistor) = 1.934834 A * 6.20 ohm = 11.9959708 V (b) What is it across each of the two sections of wire? V(wire) = 1.934834 A * 0.0010415416637355634 ohm = 0.00201521 V (c) At what rate is energy lost to thermal energy in the resistor? W = V(resistor) * I = 11.9959708 V * 1.934834 A = 23.2102122 watts (d) At what rate is it lost in each of the two sections of wire? W = V(wire) * I = 0.00389910 watts Note: on answers c and d is more appropiate to use the formula W = V^2/R, the result is the same and for simplicity in calculations I used W = V*I . For references about the concepts involved on the solution of this problem see the following pages: "Copper and electricity. Resistance and resisitivty": http://www.schoolscience.co.uk/content/5/physics/copper/copelech2pg1.html "Calculation of electrical resistivity with copper": http://www.allmeasures.com/Formulae/static/formulae/electrical_resistivity/12.htm "Ohm's Law": http://www.grc.nasa.gov/WWW/K-12/Sample_Projects/Ohms_Law/ohmslaw.html "Electric power - Wikipedia, the free encyclopedia": http://en.wikipedia.org/wiki/Electric_power Search strategy: ohm law electric power copper resistivity I hope this helps you. Regards, livioflores-ga```