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A wire is stretched so as to change its diameter by 0.25%. The percentage change in resistance is

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Important Questions on Current Electricity

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Two wires of same dimensions but resistivities ρ1 and ρ2 are connected in series. The equivalent resistivity of the combination is
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A resistance of 2 Ω is to be made from a copper wire (specific resistance=1.7×10-8 Ω m) using a wire of length 50 cm. The radius of the wire is
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The specific resistance of a wire is ρ, its volume is 3 m3 and its resistance is 3 Ω, then its length will be
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The I-V graphs for two different electrical appliances P and Q are shown in the diagram. If RP and RQ be the resistances of the devices, then

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Consider a block of conducting material of resistivity ρ  shown in the figure. Current I enters at A and leaves from D. We apply the superposition principle to find voltage Δ V  developed between B and C. The calculation is done in the following steps:
i) Take current I  entering from A and assume it to spread over a hemispherical surface in the block.
ii) Calculate field Er at distance r from A by using ohm's law E = ρ j , where r is the current per unit area at r.
iii) From the r dependence of Er, obtain the potential Er at r.
iv) Repeat (i), (ii) and (iii) for current I leaving D and superpose results for A and D.
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For current entering at A, the electric field at a distance r from A is
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Drift velocity, vd, varies with the intensity of electric field as per the relation,
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Drift speed of electrons, when 1.5 A current flows in a copper wire of cross section 5 mm2 is vd. If the electron density in copper is 9×1028 m-3 the value of vd in mm s-1 is close to (Take charge of an electron to be =1.6×10-19 C)
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A potential difference of V is applied at the ends of a copper wire of length l and diameter d. On doubling only d, the drift velocity,