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Current Electricity MCQs for NEET — Physics Questions with Answers

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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.

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Explanation

$ 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 $

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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.

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Explanation

$ r = { I_1 -I_2 \over l_2 } \times R = 0.5 \Omega $

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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

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Explanation

Total e.m.f = nE Total resistance $ R +nr \Rightarrow i = {nE \over R+nr } $

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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....

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Explanation

$ I = { E \over R+{1 \over 4}} = { 2 \over 2 + {1 \over 4 }} = { 2 \over 2.25 } = 0.888A$

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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.

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Explanation

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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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.

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Explanation

Potential gradient = $ { V \over L } = { iR \over L } = { ipl \over AL } = { ip \over A} $

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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.

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Explanation

Potential gradient x = $ { e \over ( R +Rh +r ) } {R \over L }$

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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....

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Explanation

$ E = I_1(R_1+r ) = I_2 (R_2+r)$

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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 ....

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Explanation

$ 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})]$

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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....

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Explanation

$ 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) $

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