Chapter 27, Problem 012 The figure shows a resistor of resistance R = 6.16 conne

Question

Chapter 27, Problem 012 The figure shows a resistor of resistance R = 6.16 connected to an ideal battery of emf 14.6 V by means of two copper wires. Each wire has length 22.4 cm and radius 1.00 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 (a) Number (b) Number (c) Number (d) Number Units Units Units Units Click if vou would like to Show Work for this question: Open Show Work

Explanation / Answer

Given

R = 6.16 ohm , e = 14.6 V

l = 0.224m , r =1 mm = 0.001 m

area of cross section is A = pi(0.001)^2 m^2 = 3.14159265*10^-6 m^2

the resistivity of the copper is rho = 1.68*10^-8 ohm m

we know the relation between resistivity and resistance is  

R_W = rho*l/A

R_W = (1.68*10^-8*0.224)/(3.14159265*10^-6) ohm

R_W = 0.0011978637651 ohm

now the total resistance in the circuit is  

R_t = R+2*R_w

R_t = 6.16+2*0.0011978637651 ohm = 6.1623957275302 ohm

all are in seires so the current is same across each one so

V = I*R

I = V/R = 14.6/6.1623957275302 A = 2.3692084451466 A

potential difference across the wire is V_w - I*R_w

= 2.3692084451466*0.0011978637651 V

= 0.0028379889484 V

= 2.838 mV

the potential difference across the resistor is

V_R = I*R = 2.3692084451466*6.16 V = 14.5943240 V

almost the potential difference is across the resistor so we neglect the potential difference across the wires

rate of thermal energy is P = V*I  

P_R = 2.3692084451466*6.16 W = 14.5943240 W

across the wires is  

P_W = 2.3692084451466*0.0011978637651 W = 0.002838 W

a) 14.5943240 V

b) 2.838 mV

c) 14.5943240

d) 0.002838 W

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