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Resistance of a wire at $ 50 ^\circ C $ is $5 \Omega$ , and at $ 100 ^ \circ C $ it is $6 \Omega $ find its resistance at $ 0 ^ \circ C $
$R_0 = R_0[1 + \alpha\theta]$ $5 = R_0( 1 +50 \alpha) and $ $ 6 = R_0 (1 + 100 \alpha )$
Two wires of equal dimeters of resistivities $ \rho_1 and \rho_2$ and lengths l1 and l2, respectively, are joined in series. The equivalent resistivety of the combination is
When two wires with different resistivities and lengths are connected in series, the equivalent resistivity is calculated based on the total resistance and the total length. The resistance of a wire is given by $ R = \rho \frac{l}{A} $, where \( \rho \) is the resistivity, \( l \) is the length, and \( A \) is the cross-sectional area. For wires of equal diameters, the total resistance becomes the sum of individual resistances, leading to the equivalent resistivity formula: $ \rho_{eq} = \frac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2} $.
The temperature (T) dependence of resistivity (r) of a semi-conductor is represented by
In a semiconductor when temperature increases conductivity increases so resistivity decreases i.e.
So, slope of rho - T curve is negative & it is dependent upon T i.e. it is not constant. So, alternative (3) is right choice.
Length ofa heating filament is reduced by 20% its power will....
$ P = { v^2 \over R } = { V^2 A\over \rho l} \Rightarrow P \alpha { I \over l } $ $ \therefore {P _1 \over P_2} = { l_1 \over l_2}$
What maximum power can be obtained from a battery of emf $ \varepsilon$ and internal resistance r connected with an external resistance R?
Power obtained from a battery becomes maximum when r = R $ \therefore Power P = I^2 r$ $ = ( { \varepsilon \over R+r })^2$ $r = { \varepsilon ^2 \over 4r}$
The tungsten filament of bulb has resistance equal to $18 \Omega $ at 27 °C tempreature 0.25 A of current flows, when 45V is connected to it If $ \alpha = 4.5 \times 10^{-3} K^{-1}$ for a tungsten thenfind the temperature of the filament.
$ I ={V \over R} \therefore R = 180 \Omega $ $R_\theta = R0 [ 1 + \alpha(\theta -\theta0)]$
The resistance of the wire made of silver at $ 27 ^ \circ C $ temperature is equal to $2.1 \Omega $ while at $ 100 ^ \circ C $ it is $2.7 \Omega$ calculate the temprature Coefficient of the resistivity of silver. Take the reference temperature equal to $ 20 ^ \circ C $
$R_{27} = R_{20} [(I + \alpha (27-20)] = R_{20} [1 + \alpha (7) ] $ $R_{100} = R_{20} [(I + 2(100-20)] = R_{20} [1 + \alpha (80) ] $
The temperature co-efficient of resistance of a wire is $0.00125 ^ \circ k^{-1}$. Its resistance is $1 \Omega$ at 300K. Its resistance will be $2 \Omega$ at.
$R_{300} = { 1 \times 2 \times 300 \over 1+2T } $ $ \therefore { 1 \over 2 } = { 1 + 2 \times 300 \over 1 \times 2T}$ $ \therefore 1 + \alpha T = 2 + 600 \alpha $ $ \therefore T = { 1 + 600 \alpha \over \alpha } = { 1 + (600 \times 0.00125) \over (0.00125) } = { 1.75 \over 0.00125 } =1400 K$
Two resistances $R_1 and R_ 2 $have effective resistance $R_s$ when connected in series combination and $R_p$ when connected in parallel combination if $R_s R_p= 16 and R_1/ R_ 2 = 4$ the values of $R_1 and R_ 2$ are
$R_p = { R_1 R_2 \over R_1 + R_2 }$ $Rs = R_1 +R_2 $ $ Use the \, {R_1 \over R_2 } = 4 $
Three identical resistors connected in series with a battery, together dissipate 10W of power. What will be the power dissipated, if the same resistors are connected in parallel across the same battery?
$ R_s = R_1 + R_2 + R_3 = R +R+R = 3 R $ $R_p = { 1 \over R_1} + { 1 \over R_2 } + { 1 \over R_3 } = { 1\over R }+ {1 \over R} + {1 \over R } = { 3 \over R } $ $Or R_p = { R \over 3 } $ $ P = {V^2 \over R} \alpha { 1 \over R} $ $ {P_s \over P_p} = { R_p \over R_s} Or P_p = { R_s \over R_p } \times P_s = { 3R \over R/3} \times 10 = 90 W$
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