Resistance Calculations in Electrical Conductors Quiz

<p>Spiraling</p> Signup and view all the answers

<p>$R_{dc} = \frac{\rho_T l}{A}$</p> Signup and view all the answers

<p>Current tends to crowd toward the conductor surface</p> Signup and view all the answers

<p>Cross-sectional area</p> Signup and view all the answers

<p>Resistivity increases with increasing temperature</p> Signup and view all the answers

<p>Resistance increases with increasing current magnitude</p> Signup and view all the answers

<p>$I = \frac{1}{r}$</p> Signup and view all the answers

<p>$A = 12 \pi (\frac{d}{2})^2$</p> Signup and view all the answers

<p>Resistance of ACSR conductors increases less than steel conductors with increasing current magnitude</p> Signup and view all the answers

<p>spiraling</p> Signup and view all the answers

<p>$R_{dc,T} = rac{ρ_T l}{A}$</p> Signup and view all the answers

<p>3.93 x 10^{-3} Ω·cm/°C</p> Signup and view all the answers

<p>R_{\text{dc},T} = \dfrac{\rho_T l}{A} \Omega</p> Signup and view all the answers

<p>10.37 Wm x 10^{-8} or 1.59 Ω·cm</p> Signup and view all the answers

<p>A = 12 \pi r^2 = 12 \pi \left(\dfrac{d}{2}\right)^2 = 12 \pi \left(\dfrac{3.373 \text{ mm}}{2}\right)^2 = 107.23 \text{ mm}^2</p> Signup and view all the answers

<p>1.77$ W-cmil/ft</p> Signup and view all the answers

<p>R_{\text{dc},T} = \dfrac{\rho_T l}{A} \Omega</p> Signup and view all the answers

<p>R_{\text{dc},50°C} = \left[\left(1.973 \times 10^{-8} \Omega\text{m}\right) \times \left(\dfrac{103 \times 1.02 \text{ m}}{107.23 \text{ mm}^2}\right)\right] = 0.1877 \Omega/\text{km}</p> Signup and view all the answers

<p>Table 4.3</p> Signup and view all the answers

<p>2.83$ W-cmil/ft</p> Signup and view all the answers

<p>61$ W-cmil/ft</p> Signup and view all the answers

<p>spiraling</p> Signup and view all the answers

<p>$R_{dc,T} = \frac{\rho_T l},{A}$</p> Signup and view all the answers

<p>9.6</p> Signup and view all the answers

<p>$\text{ac resistance or effective resistance} = R_{dc} \sqrt{1 + (2 \pi f \rho)/\mu}$</p> Signup and view all the answers

<p>$A = 12 \pi r^2$</p> Signup and view all the answers

<p>0.1877 Ω / km</p> Signup and view all the answers

<p>$R_{dc,50°C} = \rac{ρ_{T} l},{A}$</p> Signup and view all the answers

<p>For AC, the current distribution is non-uniform.</p> Signup and view all the answers

<p>$R_{dc,T} = \rac{ρ_{T} l},{A}$</p> Signup and view all the answers

<p>1.77 x 10^{-8} Ωm</p> Signup and view all the answers

<p>10.37 x 10^{-8} Ωm</p> Signup and view all the answers

<p>Amount of dirt, salt, and other contaminants, and meteorological factors</p> Signup and view all the answers

<p>Table 4.3</p> Signup and view all the answers

<p>Electrically ionized air due to high electric field strength at conductor surface</p> Signup and view all the answers

<p>It is a very small component of the shunt admittance</p> Signup and view all the answers

<p>High electric field strength at conductor surface causing air ionization</p> Signup and view all the answers

<p>It is a very small component of the shunt admittance</p> Signup and view all the answers

<p>Accumulation of dirt, salt, and contaminants on insulators</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>$B = \mu H$</p> Signup and view all the answers

<p>$L = \frac{\lambda}{I}$</p> Signup and view all the answers

<p>All of the above</p> Signup and view all the answers

<p>$4\pi \times 10^{-7} H/m$</p> Signup and view all the answers

<p>The inductance of each conductor due to both internal and external flux linkages is given by $Lx = Ly = 2 \times 10^{-7} \ln \left( \frac{D},{r'} \right) , \text{H/m}$, which can be converted to mH/km by multiplying by 1000.</p> Signup and view all the answers

<p>The total inductance of the line, denoted as $L$, is the sum of inductance of each conductor: $L = Lx + Ly = 0.8899 + 0.8899 = 1.780 , \text{mH/km}$ per circuit.</p> Signup and view all the answers

<p>The inductance of each conductor due to both internal and external flux linkages is given by $Lx = Ly = 2 \times 10^{-7} \ln \left( \frac{D},{r'} \right) , \text{H/m}$, where $D$ is the spacing between conductors and $r'$ is the effective radius of the conductor.</p> Signup and view all the answers

<p>The total inductance of the single-phase circuit, called loop inductance, is not provided in the given text.</p> Signup and view all the answers

<p>The equation to calculate the inductance of each conductor due to internal flux linkages only is $Lint = \frac{1}{2} \times 10^{-7} H/m = 0.05 mH/Km$ per conductor.</p> Signup and view all the answers

<p>In Example 1, the inductance of each conductor due to internal flux linkages only is 0.05 mH/Km per conductor.</p> Signup and view all the answers

<p>The flux linkage in phase a conductor for a three-phase three-wire line is not provided in the given text.</p> Signup and view all the answers

<p>0.8899 mH/Km per conductor</p> Signup and view all the answers

<p>1.780 mH/Km per circuit</p> Signup and view all the answers

<p>0.8899 mH/Km per conductor</p> Signup and view all the answers

<p>\lambda_{a}$</p> Signup and view all the answers

<p>L_{total}</p> Signup and view all the answers

<p>$L_{total} = 2 \times L_{int}$</p> Signup and view all the answers

<p>L_{int}</p> Signup and view all the answers

<p>current</p> Signup and view all the answers

<p>L</p> Signup and view all the answers

<p>La</p> Signup and view all the answers

<p>Lint = \frac{1}{2} \times 10^{-7} \frac{H}{m} = 0.05 \frac{mH}{Km} per conductor</p> Signup and view all the answers