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Temperature of a conductor increases by $5 ^\circ $C passing electric current for some time. The increase in its temperature when double current is passed through the same conductor for the same time is $ ^ \circ C $
$ \triangle Q\, \alpha\, I^2 $ V is same so $ P_1 = { V^2 \over R_1 } and P_2 = {V^2 \over R_2} $
Area of cross-section of two wires of same length carrying same current is in the ratio of 1 : 2. Then the ratio of heat generated per second in the wires = ....
$ I , \rho , l are equal so H \alpha {1 \over A} $ $ \therefore {H_1 \over H_2} = {A_2 \over A_1} = {2 \over 1} $
If $ \sigma_1, \sigma_2, \sigma_3$ are the conductances of three conductor then equivalent conductance when they are joined in series, will be.
$ Reff = R_1 =R_2 +R_3 \Rightarrow {1 \over \sigma_eff} = { 1 \over \sigma _2 } + { 1 \over \sigma _3} $
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$ $r = { \varepsilon ^2 \over 4r}$
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