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n resistors each of resistance r are connected to a battery of emf E and internal resistance r. Then the ratio of terminal voltage to emf of battery =....
$ V = \varepsilon-Ir , \varepsilon = V+Ir ,$ $ but V = nlr , \varepsilon = nlr +Ir = (n+1) Ir$ $\therefore {V \over \varepsilon} = { n \over n+1 } $
Masses of three conductors of same material are in the proportion of 1:2:3 their lengths are in the proportion of 3:2:1 then their resistance will be in the proportion of....
$Mass m = density \times volume = dAl$ $ \therefore A = {m \over dl} $ $ Now, R = { \rho l \over A} = { \rho l d l \over m} = { \rho d l^2 \over m } $ $ \therefore R \alpha {l^2 \over m } ( \therefore p and d = constant ) $ $ \therefore R_1 : R_2:R_3 = { 9 \over 1} : { 4 \over 2 } : { 1 \over 3 } = 27 :6:1 $
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$
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.