Answer to Question #283450 in Electric Circuits for kocii

Question

A 9 V battery is connected to a resistor with resistance of 1.60 Ω. (a) How many electrons leave the battery per minute? (b) If the resistor is made of copper wire with radius of 25 µm, calculate the length of the copper wire. The resistivity of copper is given as 1.68×10−8 Ω·m.

Answer 1

V=9V

R=1.60 $\Omega$

t=60sec

r=25$\mu m$

$\rho=1.68\times10^{-8}\Omega m$

Part (a)

Ohm law

V=IR

$I=\frac{V}{R}$

$I=\frac{9}{1.60}=5.625Amp$

We know that

$i=I=5.625Amp$

$nq=it\\n=\frac{it}{q}$

$q=1.6\times10^{-19}c$

Put value

$n=\frac{5.625\times60}{1.6\times10^{-19}}=2.109\times10^{21}$

Part(b)

We know that

$R=\frac{\rho l}{A}$

$$A=\pi r^2=3.14\times(25\times10^{-6})=1.9625\times10^{-9}m^2$$

$l=\frac{RA}{\rho}$

Put value

$l=\frac{1.60\times1.9625\times10^{-9}}{2.68\times10^{-8}}=0.1171 m$