Chapter 27, Problem 012 Your answer Is partially correct. Try again. The figure

Question

Chapter 27, Problem 012 Your answer Is partially correct. Try again. The figure shows a resistor of resistance R = 6.32 ? connected to an ideal battery of emf E = 12.3 V by means of two copper wires. Each wire has length 23.0 cm and radius 1.90 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? Wire Wire 2 Unitsyv (a) Numbef12.29 (b) Numberi0 (c) Numbe23.93 (d) Number 1.3e-9 Click if you would like to Show Work for this question: Open Show Work UnitsT mV UnitsT w UnitsT mw

Explanation / Answer

resistance of copper wire 1 R1 = rho*L1/A = rho*L1/(pi*r1^2)

R1 = 1.72*10^-8*0.23/(pi*(1.9*10^-3)^2)

R1 = 3.5*10^-4 ohm

resistance of copper wire 2 , R2 = rho*L2/A2 = rho*L2/(pi*r2^2)

R2 = 1.72*10^-8*0.23/(pi*(1.9*10^-3)^2)

R2 = 3.5*10^-4 ohm

total resistance of circuit Rtot = R1 + R2 + R = 6.3207 ohm

current I = E/Rtot = 1.946 A

(a)

potential difference V = I*R = 12.29 A

(b)

across each of two resistors = (E - V)/2 = (12.3-12.29)/2 = 5 mV

(c)

rate of energy loss in resistor P = I^2*R = 23.93 W

(d)

rate of energy loss in each section of wire P = I^2*R1 = 1.32 mW

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