Try the fastest way to create flashcards
Question

Can a 2.5-mm-diameter copper wire have the same resistance as a tungsten wire of the same length? Give numerical details.

Solution

Verified
Answered 2 years ago
Answered 2 years ago
Step 1
1 of 2

In order to determine the resistance we use the equation (18-3)\textbf{(18-3)}:

R=ρLA,R=\rho \frac{L}{A},

where ρ\rho represents the resistivity of a certain material, A=πr2=π(d/2)2=4πd2A=\pi r^2=\pi (d/2)^2=4\pi d^2 is a cross sectional area where dd is diameter.

In our problem we have a dc=2.5mmd_c=2.5mm copper wire with a length LL and a tungsten wire with a same length LL. Let's denote with RcR_c and RtR_t the resistances of the copper and tungsten wire respectively. If the resistances are the same then:

Rc=RtρcLAc=ρtLAtρcAt=ρtAc,\begin{equation}R_c=R_t\Rightarrow \rho_c\frac{L}{A_c}=\rho_t\frac{L}{A_t}\Rightarrow \rho_c A_t=\rho_t A_c,\end{equation}

where ρc=1.68×108Ωm\rho_c=1.68\times 10^{-8}\Omega m and ρt=5.6×108Ωm\rho_t=5.6\times 10^{-8}\Omega m (table 18-1). Therefore from (1):

ρcπdt24=ρtπdc24ρcdt2=ρtdc2.\rho_c \frac{\pi d_t^2}{4}=\rho_t \frac{\pi d_c^2}{4}\Rightarrow \rho_c d_t^2=\rho_t d_c^2.

Solve for dtd_t

dt=dcρtρc=(2.5mm)5.6×108Ωm1.68×108Ωm,d_t=d_c\sqrt{\dfrac{\rho_t}{\rho_c}}=(2.5mm)\sqrt{\dfrac{5.6\times 10^{-8}\Omega m}{1.68\times 10^{-8}\Omega m}},

dt=4.56mm.d_t=4.56mm.

The is answer is: It can if the diameter of the tungsten wire is dt=4.56mm.d_t=4.56mm.

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Physics: Principles with Applications 6th Edition by Giancoli

Physics: Principles with Applications

6th EditionISBN: 9780130606204 (13 more)Giancoli
4,385 solutions
Fundamentals of Physics 8th Edition by Halliday, Resnick, Walker

Fundamentals of Physics

8th EditionISBN: 9780471758013 (2 more)Halliday, Resnick, Walker
9,479 solutions
Fundamentals of Physics 10th Edition by David Halliday, Jearl Walker, Robert Resnick

Fundamentals of Physics

10th EditionISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick
8,971 solutions
Fundamentals of Physics, Extended 11th Edition by David Halliday, Jearl Walker, Robert Resnick

Fundamentals of Physics, Extended

11th EditionISBN: 9781119306856David Halliday, Jearl Walker, Robert Resnick
9,587 solutions

More related questions

1/4

1/7