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Current Electricity flash cards

Master Current Electricity through 90 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Current Electricity, question and answer

17 of this chapter's 90 cards, laid out open so you can read straight through. The remaining 73 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.Define electric current and give its SI unit.

    Electric current is the rate of flow of charge: I=dqdtI = \dfrac{dq}{dt}. SI unit is the ampere (A); 1A=1C/s1\,\text{A} = 1\,\text{C/s}. It is a scalar with a conventional direction (direction of positive charge flow).

    Hint: Charge per unit time.

  2. 2.What is conventional current direction versus electron flow?

    Conventional current is taken in the direction positive charges would move, i.e. from high to low potential (outside the source). Electrons actually drift opposite to this. Both describe the same physical current.

    Hint: Positive-charge convention vs real electron motion.

  3. 3.Define drift velocity and give the current relation.

    Drift velocity vdv_d is the average velocity electrons acquire along the wire under an applied field. Current I=neAvdI = neAv_d, where nn is free-electron density, ee the electronic charge, AA the cross-section.

    Hint: I=neAvdI = neAv_d.

  4. 4.Relate drift velocity to the applied electric field and relaxation time.

    vd=eEmτv_d = \dfrac{eE}{m}\tau, where τ\tau is the mean relaxation time between collisions, EE the field, mm the electron mass. Drift velocity is proportional to the field.

    Hint: Acceleration eE/meE/m acting for time τ\tau.

  5. 5.Define mobility of a charge carrier and its SI unit.

    Mobility μ=vdE=eτm\mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m}, the drift speed per unit field. SI unit is m2V1s1\text{m}^2\,\text{V}^{-1}\,\text{s}^{-1}.

    Hint: Drift speed per unit field.

  6. 6.Express current density JJ in terms of nn, vdv_d and in terms of conductivity.

    J=IA=nevdJ = \dfrac{I}{A} = nev_d. Also J=σE\vec{J} = \sigma \vec{E} (microscopic Ohm's law), where σ\sigma is conductivity.

    Hint: J=nevd=σEJ = nev_d = \sigma E.

  7. 7.State Ohm's law in its macroscopic and microscopic forms.

    Macroscopic: V=IRV = IR. Microscopic: J=σE\vec{J} = \sigma\vec{E}. Ohm's law states current is proportional to potential difference at constant temperature.

    Hint: V=IRV=IR and J=σEJ=\sigma E.

  8. 8.Define resistance and give the formula for a uniform conductor.

    Resistance R=VIR = \dfrac{V}{I}; SI unit ohm (Ω\Omega). For a uniform wire R=ρLAR = \dfrac{\rho L}{A}, with ρ\rho resistivity, LL length, AA area.

    Hint: R=ρL/AR = \rho L / A.

  9. 9.Define resistivity and relate it to microscopic quantities.

    Resistivity ρ=RAL\rho = \dfrac{RA}{L}; SI unit Ωm\Omega\,\text{m}. Microscopically ρ=mne2τ\rho = \dfrac{m}{ne^2\tau}, independent of dimensions, dependent on material and temperature.

    Hint: ρ=m/(ne2τ)\rho = m/(ne^2\tau).

  10. 10.How do resistance and resistivity change when a wire is stretched to nn times its length (volume constant)?

    Resistivity ρ\rho is unchanged (material property). Length becomes nLnL, area becomes A/nA/n, so R=ρnLA/n=n2RR' = \rho\dfrac{nL}{A/n} = n^2 R. Resistance increases n2n^2 times.

    Hint: RL2R \propto L^2 at fixed volume.

  11. 11.State the limitations/failures of Ohm's law.

    Ohm's law fails when: VV is non-linear in II (e.g. diodes), VVII relation depends on sign of VV (rectifiers), or is multivalued. Such devices are non-ohmic; also fails at high fields/temperatures.

    Hint: Non-linear, direction-dependent, or multivalued VVII.

  12. 12.Give the temperature dependence of resistivity for a conductor.

    ρT=ρ0[1+α(TT0)]\rho_T = \rho_0[1 + \alpha(T - T_0)], where α\alpha is the temperature coefficient of resistivity. For metals α>0\alpha > 0 (resistivity rises with temperature).

    Hint: ρ=ρ0(1+αΔT)\rho = \rho_0(1+\alpha\Delta T).

  13. 13.How does resistivity of metals, semiconductors, and conductors differ with temperature?

    Metals: ρ\rho increases with TT (α>0\alpha>0). Semiconductors and insulators: ρ\rho decreases with TT (α<0\alpha<0) as more carriers are freed. Alloys like nichrome/manganin have very small α\alpha.

    Hint: Metals up, semiconductors down.

  14. 14.Why does resistivity of a metal increase with temperature (microscopic reason)?

    ρ=mne2τ\rho = \dfrac{m}{ne^2\tau}. In metals nn is nearly constant, but higher temperature causes more frequent collisions, reducing relaxation time τ\tau, so ρ\rho rises.

    Hint: τ\tau falls as TT rises.

  15. 15.Define temperature coefficient of resistance α\alpha.

    α=RTR0R0(TT0)=1RdRdT\alpha = \dfrac{R_T - R_0}{R_0\,(T - T_0)} = \dfrac{1}{R}\dfrac{dR}{dT}; the fractional change in resistance per degree rise in temperature. Unit: K1\text{K}^{-1} or C1^\circ\text{C}^{-1}.

    Hint: Fractional resistance change per degree.

  16. 16.Derive the formula for resistors in series.

    Same current II flows through each; total voltage V=V1+V2+=I(R1+R2+)V = V_1+V_2+\dots = I(R_1+R_2+\dots). Hence Rs=R1+R2+R3+R_s = R_1 + R_2 + R_3 + \dots. Series resistance exceeds the largest resistor.

    Hint: Voltages add, current common.

  17. 17.Derive the formula for resistors in parallel.

    Same voltage VV across each; currents add: I=V(1R1+1R2+)I = V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dots\right). So 1Rp=1R1+1R2+\dfrac{1}{R_p} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dots. Parallel resistance is less than the smallest.

    Hint: Currents add, voltage common.

Open the interactive deck for the other 73 cards, with self-grading so the ones you keep missing come back.

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