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Current Electricity NEET Questions — Free MCQ Practice

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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

A. $ R = R_1R_2 / (R_2 - R_1)$ B. $ R = R_1R_2 / (R_1 + R_2)$ C. $ R = R_2 -R_1 $ D. $ R = R_2(R_1 + R_2) / (R_2 -R_1)$

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Explanation

$ 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 $

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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.

A. $ {P \over Q} = { R \over S_1+S_2}$ B. $ {P \over Q} = { 2R \over S_1+S_2}$ C. $ {P \over Q} = { R ( S_1+S_2) \over \rho_1 S_2}$ D. $ {P \over Q} = { R ( S_1+S_2) \over 2 S_1S_2}$

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Explanation

$ S = { S_1 S_2 \over S_1 + S_2} $ $ {P \over Q } = { R \over S}$

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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

A. $ 3 \Omega $ B. 2 C. $ 1 \Omega $ D. $ 4 \Omega $

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Explanation

$ { 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

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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.

A. 4,2 B. 2,4 C. both (a) and (b) D. neither (a) nor (b)

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Explanation

$ 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)$

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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.

A.

$ 0.5 \Omega $

B.

$ { 1 \over 3} \Omega $

C.

$ { 1 \over 4 } \Omega $

D.

$ 1 \Omega $

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Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature It at room temperature, 100w, 60w and 40w bulbs have filament resistances $R_100 , R_60 and R_ 40$ respectively the relation between these resistances is

A. $ {1 \over R_{100}} = {1 \over R_{40}} + {1 \over R_{60}}$ B. $ R_{100} = R_{40} + R_{60}$ C. $ R_{100} \gt R_{40} \gt R_{60}$ D. $ {1 \over R_{100}} \gt {1 \over R_{60}} \gt {1 \over R_{40}}$

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Explanation

The power rating of a bulb is given by P = V^2 / R, where P is the power, V is the voltage, and R is the resistance. For bulbs with different wattages at the same voltage, the resistance is inversely proportional to the power rating. Therefore, the bulb with the highest wattage will have the lowest resistance. Thus, R_100 < R_60 < R_40, which implies 1/R_100 > 1/R_60 > 1/R_40.

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Column I
(a) The unit of electrical resistivity is (b) The unit of current density is (c) The unit of electrical conductivity is (d) The unit of electric mobility is Column II (p) $ m^2S^-1V^-1$ (q) $ \Omega ^ {-1} m^{-1}$ (r.) $ Am^{-2}$ (s) $ \Omega m$

A. a - p, b - q, c - r, d - s B. a - s, b - r, c - q, d - p C. a - r, b - q, c - p, d - s D. a - q, b - r, c - s, d - p

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Explanation

The units of the given physical quantities are: (a) Electrical resistivity: Î©m, (b) Current density: A/m^2, (c) Electrical conductivity: Î©^-1m^-1, (d) Electric mobility: m^2s^-1V^-1. Matching these units to the corresponding quantities in Column II, we get: a - s, b - r, c - q, d - p.

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Match the physical quantities given in column I with their dimensional formulae given in column II -I stands for the dimension of current. Column I (a) Electromotive force (emf) (b) Resistance (c) Resistivity (d) Conductivity Column II (p) $ [ M L^2T^{3} A^{-2}]$ (q) $ [ M L^3T^{-3} A^{-2}]$ (r ) $ [ M^{-1} L^{-3}T^{3} A^{2}]$ (s) $ [ M L^2T^{-3} A^{-1}]$

A. a - s, b - p, c - q, d - r B. a - p, b - s, c - r, d - q C. a - p, b - s, c - r, d - q D. a - r, b - p, c - q, d - s

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Explanation

The dimensional formulas for the given physical quantities are: (a) Electromotive force (emf): [ML^2T^-3A^-1], (b) Resistance: [ML^2T^-3A^-2], (c) Resistivity: [ML^3T^-3A^-2], (d) Conductivity: [M^-1L^-3T^3A^2]. Matching these dimensional formulas to the quantities in Column II, we get: a - s, b - p, c - q, d - r.

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Assertion and reason type question: Assertion and reason are given in following questions each question has four options one of them is correct select it. Assertion: There is no current in the metals in the absence of electric field. Reason: Motion of free electrons is random.

A. Both assertion and reason are true and the reason is correct explanation of the assertion. B. Both assertion and reason are true, but reason is not correct explanation of the assertion. C. Assertion is true, but the reason is false. D. Both, assertion and reason are false

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Explanation

In metals, the motion of free electrons is indeed random in the absence of an electric field, which means that there is no net current. The assertion states that there is no current in metals without an electric field, and the reason given is that the motion of free electrons is random. Both statements are true, and the reason correctly explains the assertion.

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The potential difference between points A and B is -

A.

2 V

B.

6 V

C.

4 V

D.

3 V

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Explanation

Current in loop = 3 A

$V_A - V_B$ = 3 + 2 - (1 ohm) (3A) = 2 V

Current Electricity MCQs for NEET — Physics Questions with Answers

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