Current Electricity Concept Page - 5

Definition

Describe cryogenics

Cryogenics is the branch of physics that deals with the production and effects of very low temperatures. The Large Hadron Collider (LHC) is the largest cryogenic system in the world and one of the coldest places on Earth.

Example

Using Ohm's Law

Given a current of 2 A$2A$ flowing through a wire of resistance 0.1Ω$0.1\mathrm{\Omega }$. 
Then according to Ohm's Law, the potential difference across the ends of the wire is:
V=IR=2×0.1=0.2V$V=IR=2\times 0.1=0.2V$

Example

Resistance of a different geometric configuartion of wire

A variable cross section resistor treated as a serial combination of small straight wire resistors
r(x)=b+a−bhx$r(x)=b+{\displaystyle \frac{a-b}{h}}x$
and dx=ha−bdr$dx={\displaystyle \frac{h}{a-b}}dr$
R=∫dR=∫h0ρdxπr2(x)$R=\int dR={\int }_0^h\rho {\displaystyle \frac{dx}{\pi r^2(x)}}$
=∫abρhdrπr2(a−b)=ρhπab$={\int }_b^a{\displaystyle \frac{\rho hdr}{\pi r^2(a-b)}}={\displaystyle \frac{\rho h}{\pi ab}}$

Example

Current across varying resistivity

A rod of length L$L$ and cross-section area A$A$ lies along the x-axis between x=0$x=0$ and x=L$x=L$. The material obeys Ohm's law and its resistivity varies along the rod according to ρ(x)=ρ0e−x/L$\rho (x)={\rho }_0{e}^{-x/L}$. The end of the rod at x=0$x=0$ is at a potential V0$V_0$ and it is zero at x=L$x=L$Total resistance=R=∫L01Aρ(x)dx=∫L01Aρ0e−x/Ldx$\text{Total resistance}=R={\int }_0^L{\displaystyle \frac{1}{A}}\rho (x)dx={\int }_0^L{\displaystyle \frac{1}{A}}{\rho }_0{e}^{-x/L}dx$R=ρ0LA(1−1e)$R={\displaystyle \frac{{\rho }_{0}L}{A}(1-{\displaystyle \frac{1}{e}})}$
Current = V/R; I=V0Aρ0L(ee−1)$I={\displaystyle {\displaystyle \frac{V_{0}A}{{\rho }_{0}L}}\left({\displaystyle {\displaystyle \frac{e}{e-1}}}\right)}$

Formula

Procedure to calculate resistance of a resistor with varying cross sectional area

• We know resistance R=ρlA$R={\displaystyle \frac{\rho l}{A}}$
• Let us take a  resistor of varying cross-section area (area changes as we move from one end to another) as shown in the figure.
• First, take an element of dx$dx$ thickness  at length x$x$ and find the area of cross-section as a function of x A(x)$A(x)$.
• Then integrate from one end to another to calculate the resistance. 
• Hence R=∫l0ρdxA(x)$R={\int }_0^l{\displaystyle \frac{\rho dx}{A(x)}}$

Example

Example

A variable cross section resistor treated as a serial combination of small straight wire resistorsr(x)=b+a−bhx$r(x)=b+{\displaystyle \frac{a-b}{h}}x$
and dx=ha−bdr$dx={\displaystyle \frac{h}{a-b}}dr$
R=∫dR=∫h0ρdxπr2(x)$R=\int dR={\int }_0^h\rho {\displaystyle \frac{dx}{\pi r^2(x)}}$=∫abρhdrπr2(a−b)=ρhπab$={\int }_b^a{\displaystyle \frac{\rho hdr}{\pi r^2(a-b)}}={\displaystyle \frac{\rho h}{\pi ab}}$

Definition

Infinite resistors in series and parallel

The equivalent resistance between the terminals P and Q of infinite network as shown in the figure is :Let the total Resistance be X$X$.
Then we can write circuit as drawn

Reff=X$R_{eff}=X$

X=2R+RXR+X$X=2R+{\displaystyle \frac{RX}{R+X}}$

(R+X)X=2R2+2RX+RX$(R+X)X=2R^2+2RX+RX$

RX+X2=2R2+2RX+RX$RX+X^2=2R^2+2RX+RX$

X2−2RX−2R2=0$X^2-2RX-2R^2=0$

X=2R±4R2+8R2−−−−−−−−−√2$X={\displaystyle \frac{2R\pm \sqrt{4R^2+8R^2}}{2}}$

X=2R+2R3√2$X={\displaystyle \frac{2R+2R\sqrt{3}}{2}}$

     =R(1+3√)$=R(1+\sqrt{3})$ 

Definition

Resistors in series

For resistors R1,R2,....Rn$R_1,R_2,....R_n$ in series:
Req=R1+R2+....+Rn$R_{eq}=R_1+R_2+....+R_n$
Ieq=I1=I2=....=In=I$I_{eq}=I_1=I_2=....=I_n=I$
Veq=V1+V2+....Vn$V_{eq}=V_1+V_2+....V_n$

Definition

Resistors in parallel

For n$n$ resistors in parallel:
Veq=V1=V2=....=Vn$V_{eq}=V_1=V_2=....=V_n$
Ieq=I1+I2+....+In$I_{eq}=I_1+I_2+....+I_n$ 
1Req=1R1+1R2+....+1Rn

BookMarks

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