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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$
A potentiometer wire of length 1 m and resistance $10 \Omega$ is connected in series with a cell of e.m.f 2V with internal resistance $ 1 \Omega $ and a resistance box of a resistance R if potential difference between ends of the wire is 1V the value of R is.
$ I = { E \over x + R+r} = { 2 \over 10+R+1} = { 2 \over 11+R} $ $ \Rightarrow V = I \times X $ $ Or I = { 2 \over 11 + R } \times 10 = { 20 \over 11+R} $ $ Or 11 + R = 20 $ $ R = 20 -11 = 9 \Omega $
For a cellof e.m.f 2V, a balance is obtained for 50 cmof the potentiometer wire If the cell is shunted by a $2 \Omega$ resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is.
$ r = { I_1 -I_2 \over l_2 } \times R = 0.5 \Omega $
n identical cells each of e.m.f E and internal resistance r are connected in series An external resistance R is connected in series to this combination the current through R is
Total e.m.f = nE Total resistance $ R +nr \Rightarrow i = {nE \over R+nr } $
4 cell each of emf 2v and internal resistance of $1 \Omega $ are connected in parallel to a load resistor of $ 2 \Omega$ Then the current through the load resistor is....
$ I = { E \over R+{1 \over 4}} = { 2 \over 2 + {1 \over 4 }} = { 2 \over 2.25 } = 0.888A$
A Potentiometer wire, 10mlong, has a resistance of $40 \Omega $ It is connected in series with a resistance box and a 2V storage cell If the potential gradient along the wire is 0.1mv/cm, the resistance unplugged in the box is.
Potential gardient along wire = Potential difference along wire length of wire $ = 0.1 \times 10^ {-3} = {I \times 40 \over 1000} V/cm $ $ \Rightarrow current in wire , I = { 1 /400 A}$ $ {2 \over 40+R} = { 1 \over 400} or R = 800 -40 = 760 \Omega $
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