Learn definition and concept of resistance, practice current and resistance equation, and calculate resistance of truncated cone, hollow sphere & cylinder and resistance of a rod whose length varies linearly as a function.
For face (1) or (4)
Length = \(\ell\) , Area = bh
For Face (2) or face (5)
Length = b, Area = \(\ell h\)
For Face (3) or face (6)
Length = h, Area = \(b\;\ell \)
A 20 \(\Omega\;m\)
B 15 \(\Omega\;m\)
C 30 \(\Omega\;m\)
D 40 \(\Omega\;m\)
The resistivity of material is \(\rho\).
\(dR=\dfrac {\rho\,dr}{2\pi\,rL}\)
Integrating both sides,
or, \(\int\limits_0^RdR=\displaystyle\int\limits_a^b\dfrac {\rho\,dr}{2\pi\, r\,L}\)
or, \(R=\dfrac {\rho}{2\pi\, L}\displaystyle\int\limits_a^b\dfrac {1}{r}\,dr\)
\(R=\dfrac {\rho}{2\pi\, L}\ell n \left (\dfrac {b}{a} \right)\)
A \(0.5\) \(\ell n\,(2)\,\Omega\)
B \(\ell n\,(2)\,\Omega\)
C \(3\) \(\ell n\,(2)\,\Omega\)
D \(8\) \(\ell n\,(2)\,\Omega\)
Consider a hollow cylinder of inner radius \(r_1\) and outer radius \(r_2\) of length \(L\) as shown in figure.
The area of hollow cylinder = Area of shaded region
\(A=\pi\,\left(r_2^2-r_1^2\right)\)
Resistance, \(R=\dfrac{\rho L}{A}=\dfrac{\rho L}{\pi\,\left(r_2^2-r_1^2\right)}\,\)
A \(R=\dfrac {\rho\;L}{\pi(b^2-a^2)}\)
B \(R=\dfrac {\rho\;L}{\pi\; b^2}\)
C \(R=\dfrac {\rho\;L}{\pi\; a^2}\)
D \(R=\dfrac {\rho(b^2-a^2)}{L}\)
\(\rho=\rho_0+ax\)
where,
\(\rho_0\) and \(a\) are constant
\(\rho_{(x=0)}=\rho_0\)
\(\rho_{(x=\ell)}=\rho_0+a\ell\)
\(\rho_{avg}=\dfrac {\rho_{(x=0)}+\rho_{(x=\ell)}}{2}\)
\(\rho_{avg}=\dfrac {\rho_0+\rho_0+a\ell}{2}\)
\(\rho_{avg}=\rho_0+\dfrac {a\ell}{2}\)
\(R=\dfrac {\rho_{avg}\cdot \ell}{A}\)
or, \(R=\dfrac { \left ( \rho_0+\dfrac {a\ell}{2} \right)\ell} {A}\)
or, \(R=\dfrac {2\rho_0\ell+a\ell^2} {2A}\)
A 30 \(\Omega\)
B 20 \(\Omega\)
C 21 \(\Omega\)
D 24.16 \(\Omega\)
\(dR=\dfrac {\rho\;dx}{4\pi\,x^2}\) \(\left [ \therefore R=\dfrac {\rho\,L}{A} \right ]\)
Integrating both sides,
\(\int dR=\displaystyle \int \dfrac {\rho\,dx}{4\pi\,x^2}\)
or, \(\int\limits_0^R dR=\displaystyle\int\limits_a^b \dfrac {\rho\,dx}{4\pi\,x^2}\)
or, \(R=\dfrac {\rho}{4\pi}\displaystyle\int\limits_a^b \dfrac {1}{x^2}dx\)
or, \(R=\dfrac {\rho}{4\pi} \left [ -\dfrac {1}{x} \right]_a^b \)
or, \(R=\dfrac {\rho}{4\pi} \left [ \dfrac {1}{a} -\dfrac {1}{b} \right]\)
\(R=\dfrac {\rho}{4\pi} \left [ \dfrac {b-a}{ab} \right]\)
A 0.25 \(\Omega\)
B 5 \(\Omega\)
C 1 \(\Omega\)
D 2 \(\Omega\)
Truncated Cone: The truncated cone is a solid, similar to cylinder except that the circular and planes are not of same size.
\(tan\,\theta=\dfrac {b-r}{x}\) ...(i)
\(tan\,\theta=\dfrac {b-a}{h}\) ...(ii)
\(From\, (i) \,and \,(ii)\)
\(\dfrac {b-r}{x}=\dfrac {b-a}{h}\)
or, \(r=(a-b)\dfrac {x}{h}+b\)
\(dR=\dfrac {\rho\, dx}{\pi\;r^2}\)
Integrating both sides,
\(\int\limits_0^R{dR}=\displaystyle\int\limits_0^h \dfrac {\rho\;dx} {\pi \left [ (a-b)\dfrac {x}{h}+b \right]^2} \)
Let \(\dfrac {a-b}{h}=\alpha,\; b=\beta\) ...(iii)
\(R=\dfrac {\rho}{\pi} \left [ \;\displaystyle\int\limits_0^h\dfrac {dx}{(\alpha x+\beta)^2} \;\right]\)
\(\therefore\) As we know, \(\displaystyle\int\dfrac {dx}{(\alpha x+\beta)^2} =\dfrac {-1}{\alpha(\alpha x+\beta)}\)
\(R=\dfrac {\rho}{\pi} \left [ \dfrac {-1}{\alpha(\alpha x + \beta)} \right]_0^h\)
or, \(R=\dfrac {\rho}{\pi} \left [ \dfrac {-1}{\alpha(\alpha h + \beta)} +\dfrac {1}{\alpha \beta}\right]\)
or, \(R=\dfrac {\rho}{\pi} \left [ \dfrac {-\beta+\alpha h+\beta}{\alpha(\alpha h + \beta)\beta} \right ]\)
or, \(R=\dfrac {\rho}{\pi} \left [ \dfrac {h}{(\alpha h + \beta)\beta} \right ]\)
\(From\, equation\, (iii)\)
\(R=\dfrac {\rho\,h}{\pi} \left [ \dfrac {1}{[a-b+b]b} \right ]\)
\(R=\dfrac {\rho\,h}{\pi\;ab}\)
A 1 \(\Omega\)
B 3 \(\Omega\)
C 2 \(\Omega\)
D 4 \(\Omega\)