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The resistivity of a potentiometer wire is $ 40 \times 10 ^ {-8} ohm m$ and its area of cross-section is $ 8 \times 10 ^ { -6 } m^2 $. If 0.2 amp current is flowing through the wire, the potential gradient will be.
Potential gradient = $ { V \over L } = { iR \over L } = { ipl \over AL } = { ip \over A} $
Potentiometer wire of length 1mis connected in series with $490 \Omega $ resistance and 2V battery If 0.2 mv/ cm is the potential gradient, then resistance of the potentiometer wire is.
Potential gradient x = $ { e \over ( R +Rh +r ) } {R \over L }$
A cell supplies a current I, through aresistance$ R_1$ and a current$ I_2$ through a resistance $R_ 2$ the internal resistance of a cell is....
$ E = I_1(R_1+r ) = I_2 (R_2+r)$
Two wires of resistances R1 and R 2 have temperature coeffcient of resistances $ \alpha _1 $ and $ \alpha_2 $ respectively they are joined in series the effective temperature coefficient of resistance is ....
$ R_it = R_1 ( 1 + \alpha it ) R_2 = R_2 ( 1 + \alpha_2 t )$ $ R_st = R_1 t + R_2 t $ $ = R_1[ 1 + \alpha_1 t ] + R_2[ 1 + \alpha_2 t ] $ $ = R_s [ 1+ ({(R_1 \alpha_1 +R_1 \alpha_2 t )\over R_1+R_2})]$
The resistance of the series combination of two resistances is S, when they are joined in parallel the total resistance is P If S=n P, then the minimum possible value of n is....
$ S = R_1 + R_2 ; P = { R_1R_2 \over R_1 +R_2 } ; S = nP$ $ R_1 + R_2 = { nR_1R_2 \over R_1 + R_2 } $ or $ R_1^2 + R_2^2+2R_1R_2 = nR_1R_2$ or $(R_1 -R_2) ^2 + 4R_1R_2 = nR_1R_2$ or $ (R_1 -R_2)^2 -R_1R_2 (n-4)$ $ If R_1 = R_2 , Then (n-4) = 0 or (n=4) $
Two sources of equal emf are connected to an external resistance R the internal resistance of the two soureces are $R_1 and R_2 ( R_ 2 > R_1 )$ if the potential difference across the source having internalresistance R 2 is Zero, then
$ I = { E+E \over R_1 + R_2 +R } $ $ R_2 = { 2E \times R_2 \over (R_1 + R_2 + R)}$ $ \Rightarrow E - { 2ER_2 \over R_1 + R _2 +R } = 0 or E = { 2ER_2 \over R_1 + R_2 +R} $ $ R = R_2 - R_1 $
In a wheatstone's bridge, three resistance P, Q and R connected in three are a and the fourth arm is formed by two resistances $S_1 and S_2$ connected in paralled The condition for bridge to be balanced will be.
$ S = { S_1 S_2 \over S_1 + S_2} $ $ {P \over Q } = { R \over S}$
The resistance of a wire is $5 \Omega $ at $ 50 ^\circ C $ and $6 \Omega $ at $ 100 ^\circ C $ The resistance of the wire at $ 0 ^ \circ C $ will be
$ { P \over 1/3 } = { Q \over 1 - (1/3)} , 1 =1 m$ or 3P = 3/2 Q or P = Q/2 $ { P+6 \over 2/3 } = { Q \over 1/3 } $ $ P+6 = 2 Q $ $ 6 = 2 Q - { Q \over 2 } = { 3Q \over 2 } $ So Q = 4 and P =2
Resistors P and Q connected in the gaps of the meter bridge. the balancing point is obtained 1/3 m from the zero end If a $6 \Omega $ resistance is connected in series with p the balance point shifts to 2/3m form same end P and Q are.
$ R = R_0 [1 + 2t ] $ $ for series connection R_s = R_1 + R_2 $ $ at 0 ^ \circ C tap R_s = R_o + R_0 + 2 R_0 $ $2R_0 [ 1 + \alpha st ] = R_0 [ 1 + \alpha_1 t ] + R_0[1 + \alpha _2 t ]$ $ \therefore \alpha _3 = { 1 \over \alpha } (\alpha_1 + \alpha_2) ...(1) $ $ for parallel connection { 1 \over R_p} = {1 \over R_1} + {1 \over R_2}$ $ at 0 ^\circ C temp RP = R_0 /2 $ $ \therefore { 1 \over R_0/2(1 + \alpha_p t ) }= { 1 \over R_0[ 1 + \alpha_1 t]} + { 1 \over R_0[1+\alpha_2 t ]} $ $ \therefore \alpha_p = {1 \over \alpha} (\alpha_1 + \alpha _2)$
2 A current is obtained when a $2 \Omega $ resistor is connectd with battery having $ r \Omega $ as internal resistance 0.5A current is obtained if the above battery is connected to $ 9 \Omega $ resistor. Culculate the internal resistance of the battery.
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