Class 12 Physics MCQs | Chapter 3: Current Electricity – Part 1

Explanation: Electric current is defined with respect to a cross-section of a conductor, not with respect to the total charge present in the whole conductor. If a charge \(dq\) crosses the cross-section in time \(dt\), the current is \(I=\frac{dq}{dt}\). This definition focuses on charge flow rate, so two conductors may contain many free electrons but still have zero current if there is no net charge crossing. The speed of individual electrons is a microscopic idea, while current is the macroscopic rate of charge transfer through an area. A common mistake is to confuse “large number of electrons in a metal” with “large current”; current needs directed flow through a surface.

Explanation: \( \textbf{Given:} \) The charge-time relation is \(q=2t^2+3t\), with \(q\) in \(\text{C}\) and \(t\) in \(\text{s}\). \( \textbf{Required:} \) Instantaneous current at \(t=4\,\text{s}\) and average current from \(0\,\text{s}\) to \(4\,\text{s}\). \( \textbf{Instantaneous-current idea:} \) Instantaneous current is the rate of change of charge at that instant, so \(I=\frac{dq}{dt}\). \( \textbf{Differentiation:} \) \(\frac{dq}{dt}=\frac{d}{dt}(2t^2+3t)=4t+3\). \( \textbf{At } t=4\,\text{s}\textbf{:} \) \(I=4(4)+3=16+3=19\,\text{A}\). \( \textbf{Average-current idea:} \) Average current over a time interval is \(\frac{\Delta q}{\Delta t}\), not necessarily the same as the instantaneous current. \( \textbf{Charge at } t=4\,\text{s}\textbf{:} \) \(q(4)=2(4)^2+3(4)=32+12=44\,\text{C}\). \( \textbf{Charge at } t=0\,\text{s}\textbf{:} \) \(q(0)=0\,\text{C}\). \( \textbf{Average current:} \) \(I_{\text{avg}}=\frac{44-0}{4-0}=11\,\text{A}\). \( \textbf{Unit check:} \) \(\text{C s}^{-1}\) is equal to \(\text{A}\). \( \textbf{Final answer:} \) The instantaneous current is \(19\,\text{A}\) and the average current is \(11\,\text{A}\); the common mistake is to use \(q/t\) for the instantaneous value when \(q\) is not directly proportional to \(t\).

Explanation: In a metal, the moving charge carriers are electrons, which have negative charge. Conventional current is defined as the direction in which positive charge would flow. Because electrons are negatively charged, their drift direction is opposite to the direction of conventional current. Random thermal motion does not decide the current direction; only the small net drift caused by the electric field matters. Current is treated as scalar in circuit calculations, but we still assign a direction along a wire for sign convention and circuit tracing. A common mistake is to copy the electron-flow direction directly as the conventional-current direction.

Explanation: \( \textbf{Given:} \) Number of electrons crossing is \(N=2.5\times10^{19}\), time is \(t=4.0\,\text{s}\), and electronic charge magnitude is \(e=1.6\times10^{-19}\,\text{C}\). \( \textbf{Required:} \) Magnitude and direction of conventional current. \( \textbf{Charge magnitude:} \) The magnitude of charge carried by \(N\) electrons is \(q=Ne\). \( \textbf{Substitution:} \) \(q=(2.5\times10^{19})(1.6\times10^{-19})\,\text{C}\). \( \textbf{Power-of-ten simplification:} \) \(10^{19}\times10^{-19}=10^0=1\). \( \textbf{Charge crossed:} \) \(q=2.5\times1.6=4.0\,\text{C}\). \( \textbf{Current magnitude:} \) \(I=\frac{q}{t}=\frac{4.0\,\text{C}}{4.0\,\text{s}}=1.0\,\text{A}\). \( \textbf{Direction convention:} \) The electrons move from left to right, so conventional current is from right to left. \( \textbf{Unit check:} \) \(\frac{\text{C}}{\text{s}}=\text{A}\). \( \textbf{Final answer:} \) The conventional current is \(1.0\,\text{A}\), from right to left; the direction reversal comes only from the negative sign of electron charge.

Explanation: Electric current has a sign convention along a branch, but it is not a vector like force or electric field. In circuit theory, a current marked in a chosen branch direction can be positive or negative depending on actual flow, but it is not resolved into rectangular components. At a junction, Kirchhoff-type current counting uses algebraic inflow and outflow, which reflects charge conservation rather than vector addition. If current were treated as a vector in the circuit-theory sense, currents in differently oriented wires would need vector addition, which is not how branch currents are handled. The direction arrow on a circuit diagram is a reference direction, not proof that current is a vector quantity.

Explanation: \( \textbf{Graph meaning:} \) In a \(q\)-\(t\) description, current equals the slope of the charge-time graph. \( \textbf{First interval:} \) From \(0\,\text{s}\) to \(2\,\text{s}\), charge changes from \(0\,\text{C}\) to \(6\,\text{C}\). \( \textbf{Current during first interval:} \) \(I=\frac{\Delta q}{\Delta t}=\frac{6-0}{2-0}=3.0\,\text{A}\). \( \textbf{Second interval check:} \) From \(2\,\text{s}\) to \(5\,\text{s}\), \(q\) is constant, so the slope is \(0\) and current is \(0\,\text{A}\) during that part. \( \textbf{Whole interval:} \) The total charge crossing from \(0\,\text{s}\) to \(5\,\text{s}\) is still \(6\,\text{C}\). \( \textbf{Average current:} \) \(I_{\text{avg}}=\frac{6\,\text{C}}{5\,\text{s}}=1.2\,\text{A}\). \( \textbf{Final answer:} \) The first-interval current is \(3.0\,\text{A}\) and the whole-interval average current is \(1.2\,\text{A}\); a flat part of a \(q\)-\(t\) graph means zero current, not zero charge.

Explanation: A steady current means that the current at a given section remains constant with time. In an unbranched wire, the same amount of charge must pass each cross-section per unit time; otherwise charge would start accumulating somewhere between two sections. Continuous accumulation would change the electric field inside the conductor and the current would not remain steady. The electrons do not need to be used up or created along the wire, and the metal already contains a large number of free electrons. The key condition is steady charge flow through cross-sections, not identical microscopic paths of individual electrons.

Explanation: \( \textbf{Steady-current test:} \) A steady current requires equal charge crossing in equal time intervals. \( \textbf{Wire P:} \) The charge values are \(4\,\text{C}\), \(8\,\text{C}\), and \(12\,\text{C}\) at \(1\,\text{s}\), \(2\,\text{s}\), and \(3\,\text{s}\), so charge increases by \(4\,\text{C}\) in each \(1\,\text{s}\) interval. \( \textbf{Current in P:} \) \(I=\frac{\Delta q}{\Delta t}=\frac{4\,\text{C}}{1\,\text{s}}=4\,\text{A}\). \( \textbf{Wire Q:} \) The increases are \(6\,\text{C}\) and \(10\,\text{C}\) in successive \(1\,\text{s}\) intervals, so its current is changing. \( \textbf{Wire R:} \) The increases are \(3\,\text{C}\) and \(1\,\text{C}\) in successive \(1\,\text{s}\) intervals, so its current is also changing. \( \textbf{Final answer:} \) Wire \(P\) carries a steady current of \(4\,\text{A}\). The key check is equal charge increase in equal time intervals, not the largest listed charge value.

Explanation: For a charge-time graph, the slope represents current because \(I=\frac{dq}{dt}\). A straight line through the origin means \(q\) is directly proportional to \(t\), so the current is constant. If the slope is doubled, the rate of charge crossing is doubled. For the same time interval, a graph with twice the slope reaches twice the charge value. This is different from a voltage-current graph, where the meaning of slope depends on which quantity is on the vertical axis. A reliable graph check is to first identify the axes, then interpret the slope.

Explanation: \( \textbf{Known current at section } P\textbf{:} \) \(3\,\text{C}\) crosses section \(P\) in \(2\,\text{s}\). \( \textbf{Current calculation:} \) \(I=\frac{q}{t}=\frac{3\,\text{C}}{2\,\text{s}}=1.5\,\text{A}\). \( \textbf{Steady unbranched condition:} \) In a steady current through an unbranched wire, the same current crosses every cross-section. \( \textbf{Current at } R\textbf{:} \) Therefore the current at section \(R\) is also \(1.5\,\text{A}\). \( \textbf{Required charge at } R\textbf{:} \) Use \(q=It\) for \(t=6\,\text{s}\). \( \textbf{Substitution:} \) \(q=(1.5\,\text{A})(6\,\text{s})=9\,\text{C}\). \( \textbf{Unit check:} \) Since \(\text{A}=\text{C s}^{-1}\), multiplying by \(\text{s}\) gives \(\text{C}\). \( \textbf{Final answer:} \) \(9\,\text{C}\) crosses section \(R\) in \(6\,\text{s}\); the position of the cross-section does not change the steady current in an unbranched wire.

Explanation: The expression \(I=\frac{dq}{dt}\) gives the instantaneous rate at which charge crosses a chosen section. The unit \(1\,\text{A}\) means \(1\,\text{C}\) of charge crossing per \(1\,\text{s}\), so \(1\,\text{A}=1\,\text{C s}^{-1}\). Conventional current is defined in the direction in which positive charge would flow, even though electrons in metals drift oppositely. A steady current is one whose value does not change with time. The common matching error is to associate \(I=\frac{dq}{dt}\) with only average current; for average current over a finite interval, the appropriate form is \(I_{\text{avg}}=\frac{\Delta q}{\Delta t}\).

Explanation: Conventional current follows the direction of positive charge flow. In a metallic conductor, the mobile charge carriers are electrons, so their drift direction is opposite to the conventional current direction. Calling current a scalar in elementary circuit theory means it is not added by vector rules, not that direction labels are useless. Direction labels in circuit diagrams help decide signs for current, potential drops, and later circuit equations. The magnitude of current depends on charge per unit time, so slower charge crossing reduces current if the charge amount is fixed. The tempting mistake is to mix the scalar nature of current with the absence of a reference direction.

Explanation: Current density measures how much current flows through unit cross-sectional area. If the current is uniformly distributed and is normal to the cross-section, the magnitude is \(J=\frac{I}{A}\). This is different from total current \(I\), because two wires may carry the same \(I\) but have different \(J\) if their areas are different. A smaller cross-sectional area gives a larger current density for the same current. The important condition is that the current must be uniformly distributed and perpendicular to the chosen area; otherwise the simple form \(J=\frac{I}{A}\) needs vector-area treatment.

Explanation: \( \textbf{Given:} \) Current \(I=2.0\,\text{A}\) and radius \(r=0.50\,\text{mm}=0.50\times10^{-3}\,\text{m}=5.0\times10^{-4}\,\text{m}\). \( \textbf{Required:} \) Current density \(J\). \( \textbf{Area relation:} \) For a circular cross-section, \(A=\pi r^2\). \( \textbf{Substitution in area:} \) \(A=3.14(5.0\times10^{-4})^2\,\text{m}^2\). \( \textbf{Squaring the radius:} \) \((5.0\times10^{-4})^2=25\times10^{-8}=2.5\times10^{-7}\). \( \textbf{Area value:} \) \(A=3.14\times2.5\times10^{-7}=7.85\times10^{-7}\,\text{m}^2\). \( \textbf{Current-density formula:} \) \(J=\frac{I}{A}\), because the current is uniform and normal to the cross-section. \( \textbf{Substitution:} \) \(J=\frac{2.0}{7.85\times10^{-7}}\,\text{A m}^{-2}\). \( \textbf{Simplification:} \) \(J=0.255\times10^7\,\text{A m}^{-2}=2.55\times10^6\,\text{A m}^{-2}\). \( \textbf{Unit check:} \) \(\text{A}/\text{m}^2=\text{A m}^{-2}\). \( \textbf{Final answer:} \) \(J=2.55\times10^6\,\text{A m}^{-2}\); using diameter as radius would make the area four times too large and the current density four times too small.

Explanation: \( \textbf{Known condition:} \) Both wires carry the same current \(I\). \( \textbf{Current-density relation:} \) For uniform current normal to the cross-section, \(J=\frac{I}{A}\). \( \textbf{Area-radius relation:} \) For a circular wire, \(A=\pi r^2\). \( \textbf{Radius comparison:} \) If \(r_Q=2r_P\), then \(A_Q=\pi(2r_P)^2=4\pi r_P^2=4A_P\). \( \textbf{Using same current:} \) \(J_P=\frac{I}{A_P}\) and \(J_Q=\frac{I}{A_Q}=\frac{I}{4A_P}\). \( \textbf{Ratio:} \) \(\frac{J_Q}{J_P}=\frac{I/(4A_P)}{I/A_P}=\frac{1}{4}\). \( \textbf{Final answer:} \) \(\frac{J_Q}{J_P}=\frac{1}{4}\); the square in \(A=\pi r^2\) is the important distinction in radius-based current-density comparisons.

Explanation: Current density \(\vec{J}\) is a vector quantity, unlike branch current \(I\), which is treated algebraically in circuit theory. The direction of \(\vec{J}\) is the direction of conventional current, meaning the direction in which positive charge would flow. In a metallic conductor, electrons drift opposite to conventional current because they carry negative charge. Therefore \(\vec{J}\) is opposite to electron drift direction in metals. The common misconception is to transfer the scalar treatment of circuit current directly to \(\vec{J}\), but current density includes directional information per unit area.

Explanation: \( \textbf{Vector-area idea:} \) Current through a surface depends on the component of \(\vec{J}\) normal to the surface. \( \textbf{Area vector:} \) The area vector \(\vec{A}\) is perpendicular to the surface and has magnitude \(A\). \( \textbf{General relation:} \) The current is \(I=\vec{J}\cdot\vec{A}\). \( \textbf{Dot product:} \) If \(\theta\) is the angle between \(\vec{J}\) and the normal to the surface, then \(\vec{J}\cdot\vec{A}=JA\cos\theta\). \( \textbf{Special case check:} \) If \(\theta=0^\circ\), the flow is normal to the surface and \(I=JA\). \( \textbf{Tangential case check:} \) If \(\theta=90^\circ\), the current density is parallel to the surface and no current crosses it, so \(I=0\). \( \textbf{Final answer:} \) \(I=JA\cos\theta\); using \(\sin\theta\) usually comes from measuring the angle with the surface instead of with the surface normal.

Explanation: Current density is defined as current per unit area, so \(J=\frac{I}{A}\) for uniform normal flow. If the original current density is \(J_1=\frac{I}{A}\), the new current density is \(J_2=\frac{2I}{2A}\). Simplifying gives \(J_2=\frac{I}{A}=J_1\). The change in total current alone is not enough to decide \(J\); the area change must also be considered. A useful check is that current density describes crowding of current through area, not merely the amount of current in the wire.

Explanation: \( \textbf{Definition:} \) Current density is current per unit cross-sectional area. \( \textbf{Formula:} \) \(J=\frac{I}{A}\). \( \textbf{Unit of current:} \) The SI unit of \(I\) is \(\text{A}\). \( \textbf{Unit of area:} \) The SI unit of \(A\) is \(\text{m}^2\). \( \textbf{Combining units:} \) The unit of \(J\) is \(\frac{\text{A}}{\text{m}^2}=\text{A m}^{-2}\). \( \textbf{Dimension of current:} \) Current has dimension \([A]\). \( \textbf{Dimension of area:} \) Area has dimension \([L^2]\). \( \textbf{Dimensional formula:} \) \([J]=\frac{[A]}{[L^2]}=[AL^{-2}]\). \( \textbf{Final answer:} \) The correct unit and dimension are \(\text{A m}^{-2}\) and \([AL^{-2}]\); using \(\text{C m}^{-2}\) confuses current density with charge per area.

Explanation: Statement I is correct because electric current \(I\) is the rate of charge crossing a surface, written as \(I=\frac{dq}{dt}\). Statement II is correct because current density \(\vec{J}\) is current per unit area and points along conventional current direction. Statement III is also correct, but only under the condition of uniform current density normal to the surface. If the current density is not normal or is not uniform, the more general idea is based on adding normal components over the surface. The main distinction is that \(I\) is a total flow rate through the whole cross-section, while \(\vec{J}\) describes how that flow is distributed in space and direction.