Current Electricity MCQs with Answers – Part 3 (Class 12 Physics)

Explanation: A parallel combination provides more than one path for charge flow between the same two junctions. Adding another branch increases the total current drawn from the same potential difference. Since equivalent resistance is \(R_p=\frac{V}{I_{\text{total}}}\), a larger total current means a smaller equivalent resistance. This is why \(R_p\) becomes smaller than the smallest individual branch resistance. Direct addition of resistances applies to series circuits, not parallel circuits.

Explanation: In a parallel circuit, each branch has the same potential difference \(V\). The current in a branch is given by \(I=\frac{V}{R}\). For the \(R\) branch, \(I_R=\frac{V}{R}\). For the \(2R\) branch, \(I_{2R}=\frac{V}{2R}\). Therefore, \(I_R=2I_{2R}\). In parallel circuits, the smaller resistance branch carries the larger current.

Explanation: \( \textbf{Parallel voltage condition:} \) Each resistor has \(12\,\text{V}\) across it. \( \textbf{Current through \(4\,\Omega\):} \) \[ I_1=\frac{12}{4}=3\,\text{A} \] \( \textbf{Current through \(6\,\Omega\):} \) \[ I_2=\frac{12}{6}=2\,\text{A} \] \( \textbf{Current through \(12\,\Omega\):} \) \[ I_3=\frac{12}{12}=1\,\text{A} \] \( \textbf{Total current in parallel:} \) \[ I=I_1+I_2+I_3 \] \( \textbf{Substitution:} \) \[ I=3+2+1=6\,\text{A} \] The branch currents add because the source current splits at the junctions. \( \textbf{Final answer:} \) The total current drawn is \(6\,\text{A}\).

Explanation: In a parallel circuit, both resistors have the same voltage \(V\). The branch current through \(R_1\) is \(I_1=\frac{V}{R_1}\), and through \(R_2\) it is \(I_2=\frac{V}{R_2}\). The total current is \(I=I_1+I_2\). For two parallel resistors, the current divides inversely in the ratio of resistances. Therefore, the current through \(R_1\) is \(I_1=I\frac{R_2}{R_1+R_2}\). The smaller resistance branch receives the larger current, so the other branch resistance appears in the numerator.

Explanation: \( \textbf{Given resistors:} \) \(R_1=2\,\Omega\) and \(R_2=6\,\Omega\). \( \textbf{Total current:} \) \(I=8\,\text{A}\). \( \textbf{Current division for \(2\,\Omega\):} \) \[ I_1=I\frac{R_2}{R_1+R_2} \] \( \textbf{Substitution:} \) \[ I_1=8\frac{6}{2+6} \] \( \textbf{Simplification:} \) \[ I_1=8\frac{6}{8} \] \( \textbf{Calculation:} \) \[ I_1=6\,\text{A} \] The smaller resistor carries the larger branch current in a parallel connection. \( \textbf{Final answer:} \) The current through the \(2\,\Omega\) resistor is \(6\,\text{A}\).

Explanation: For \(n\) identical resistors each of resistance \(R\) in parallel, the reciprocal relation is used. Here \(n=4\), so \[ \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \] \( \textbf{Add the terms:} \) \[ \frac{1}{R_p}=\frac{4}{R} \] \( \textbf{Invert both sides:} \) \[ R_p=\frac{R}{4} \] Parallel branches provide four equal paths for current, so the equivalent resistance becomes one-fourth of each branch resistance. \( \textbf{Final answer:} \) The equivalent resistance is \(\frac{R}{4}\).

Explanation: In a parallel combination, the branch with very large resistance carries very small current. The low-resistance branch dominates the total current drawn from the source. Since the total current is nearly the same as the current through the \(10\,\Omega\) branch, the equivalent resistance is close to \(10\,\Omega\). It is slightly less than \(10\,\Omega\), not greater than it. Adding resistances directly would apply only to a series combination.

Explanation: Conductance \(G\) is defined by \(G=\frac{1}{R}\). For parallel resistors, the equivalent resistance satisfies \(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots\). Rewriting each reciprocal as conductance gives \(G_p=G_1+G_2+\cdots\). This is why parallel combinations are sometimes described as conductance addition. Extra parallel branches increase total conductance and reduce equivalent resistance.

Explanation: Row P is suitable because all parallel branches are connected between the same two junctions. Row Q is suitable because the source current divides into branch currents and then recombines. Row S is suitable because conductances add in parallel. Row R is unsuitable because the equivalent resistance of a parallel combination is smaller than the smallest branch resistance. The “larger than largest” rule belongs to series combinations.

Explanation: Adding a resistor in parallel creates an additional path between the same two junctions. For the same applied voltage, the extra branch draws additional current. Since the total current increases, the equivalent resistance \(R_p=\frac{V}{I_{\text{total}}}\) decreases. The reciprocal relation also shows this because \(\frac{1}{R_p}\) increases when a positive term \(\frac{1}{R}\) is added. The Reason gives the physical explanation behind the reduction in equivalent resistance.

Explanation: \( \textbf{Series equivalent resistance:} \) \[ R_s=4+12=16\,\Omega \] \( \textbf{Parallel equivalent resistance:} \) \[ R_p=\frac{R_1R_2}{R_1+R_2} \] \( \textbf{Substitution:} \) \[ R_p=\frac{(4)(12)}{4+12} \] \( \textbf{Calculation:} \) \[ R_p=\frac{48}{16}=3\,\Omega \] \( \textbf{Required ratio:} \) \[ R_s:R_p=16:3 \] The same two resistors can give very different equivalents depending on whether the current has one path or multiple paths. \( \textbf{Final answer:} \) The ratio is \(16:3\).

Explanation: \( \textbf{Parallel part:} \) The \(3\,\Omega\) and \(6\,\Omega\) resistors are in parallel. \[ \frac{1}{R_p}=\frac{1}{3}+\frac{1}{6} \] \( \textbf{Common denominator:} \) \[ \frac{1}{R_p}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2} \] \( \textbf{Parallel equivalent:} \) \[ R_p=2\,\Omega \] \( \textbf{Series addition:} \) This \(2\,\Omega\) is in series with \(6\,\Omega\). \[ R_{\text{eq}}=6+2=8\,\Omega \] \( \textbf{Source voltage:} \) \(V=16\,\text{V}\). \( \textbf{Total current:} \) \[ I=\frac{V}{R_{\text{eq}}}=\frac{16}{8}=2\,\text{A} \] The parallel group must be reduced before adding the series resistor. \( \textbf{Final answer:} \) The source current is \(2\,\text{A}\).

Explanation: A mixed network may contain both series and parallel parts. Resistors are in series only when the same current must pass through them without branching between them. Resistors are in parallel only when they are connected between the same two junctions and therefore have the same potential difference. Adding all resistances directly would be valid only for a pure series circuit. A correct reduction begins by recognizing the local connection pattern.

Explanation: \( \textbf{Series branch:} \) The \(2\,\Omega\) and \(4\,\Omega\) resistors are in series. \[ R_{\text{branch}}=2+4=6\,\Omega \] \( \textbf{Parallel part:} \) This \(6\,\Omega\) branch is in parallel with \(3\,\Omega\). \[ \frac{1}{R_{\text{eq}}}=\frac{1}{6}+\frac{1}{3} \] \( \textbf{Common denominator:} \) \[ \frac{1}{R_{\text{eq}}}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2} \] \( \textbf{Equivalent resistance:} \) \[ R_{\text{eq}}=2\,\Omega \] The \(2\,\Omega\) and \(4\,\Omega\) resistors are not separately parallel with \(3\,\Omega\); their series branch must be reduced first. \( \textbf{Final answer:} \) The equivalent resistance is \(2\,\Omega\).

Explanation: \( \textbf{Parallel branch condition:} \) Each parallel branch has the full source voltage. \( \textbf{Voltage across \(12\,\Omega\) branch:} \) \[ V=24\,\text{V} \] \( \textbf{Branch current relation:} \) \[ I=\frac{V}{R} \] \( \textbf{Substitution for \(12\,\Omega\) branch:} \) \[ I=\frac{24}{12} \] \( \textbf{Calculation:} \) \[ I=2\,\text{A} \] The \(4\,\Omega\) and \(8\,\Omega\) series branch affects total source current, but it does not reduce the voltage across the \(12\,\Omega\) branch. \( \textbf{Final answer:} \) The current through the \(12\,\Omega\) branch is \(2\,\text{A}\).

Explanation: \( \textbf{Branch P:} \) The \(5\,\Omega\) and \(10\,\Omega\) resistors are in series. \[ R_P=5+10=15\,\Omega \] \( \textbf{Branch Q:} \) \[ R_Q=15\,\Omega \] \( \textbf{Parallel combination:} \) Two equal \(15\,\Omega\) branches are in parallel. \[ \frac{1}{R_{\text{eq}}}=\frac{1}{15}+\frac{1}{15} \] \( \textbf{Add reciprocals:} \) \[ \frac{1}{R_{\text{eq}}}=\frac{2}{15} \] \( \textbf{Equivalent resistance:} \) \[ R_{\text{eq}}=\frac{15}{2}=7.5\,\Omega \] Two equal resistances in parallel give half of either resistance. \( \textbf{Final answer:} \) The equivalent resistance is \(7.5\,\Omega\).

Explanation: Parallel connection is a junction-based idea, not a drawing-style idea. Two resistors are in parallel when both ends of one resistor connect to the same two nodes as both ends of the other resistor. Then the potential difference across them is the same. Physical closeness or side-by-side drawing does not guarantee this condition. In circuit reduction, the junctions must be traced before deciding whether resistors are in series or parallel.

Explanation: \( \textbf{Parallel network:} \) Two \(8\,\Omega\) resistors in parallel have equivalent resistance \[ R_p=\frac{8}{2}=4\,\Omega \] \( \textbf{Series with external resistor:} \) This \(4\,\Omega\) equivalent is in series with another \(4\,\Omega\) resistor. \[ R_{\text{total}}=4+4=8\,\Omega \] \( \textbf{Source current:} \) \[ I=\frac{16}{8}=2\,\text{A} \] \( \textbf{Voltage across parallel network:} \) \[ V_p=IR_p=(2)(4)=8\,\text{V} \] The two equal series parts share the source voltage equally after the parallel pair is replaced by its equivalent resistance. \( \textbf{Final answer:} \) The voltage across the parallel network is \(8\,\text{V}\).

Explanation: Statement I is true because resistances in a true series group add directly. Statement II is true because resistances in a true parallel group obey the reciprocal relation. Statement III is false because not every network is reducible by simple series-parallel steps. A bridge-like network may contain resistors that are neither purely series nor purely parallel. Such a network needs a bridge condition or a more general circuit method before it can be simplified safely.

Explanation: \( \textbf{Series part:} \) The \(3\,\Omega\) resistor carries the total current before the current splits. \( \textbf{Current entering parallel group:} \) \[ I=3\,\text{A} \] \( \textbf{Parallel branch resistances:} \) \(R_1=6\,\Omega\) and \(R_2=12\,\Omega\). \( \textbf{Current division through \(6\,\Omega\):} \) \[ I_1=I\frac{R_2}{R_1+R_2} \] \( \textbf{Substitution:} \) \[ I_1=3\frac{12}{6+12} \] \( \textbf{Simplification:} \) \[ I_1=3\frac{12}{18}=3\left(\frac{2}{3}\right) \] \( \textbf{Branch current:} \) \[ I_1=2\,\text{A} \] The series resistor affects the total current entering the parallel group, but the branch split is controlled by the two parallel resistances. \( \textbf{Final answer:} \) The current through the \(6\,\Omega\) branch is \(2\,\text{A}\).