Current Electricity – NEET UG PhysicsPractice Questions & PYQs

The electrical resistance of a uniform wire is directly proportional to the square of its length and inversely proportional to its mass, expressed as $R \propto \frac{l^2}{m}$ since mass accounts for the product of length, cross-sectional area, and density. Using this proportional relationship, the resistance ratio for the three wires is found by evaluating $\frac{l_1^2}{m_1}$ to $\frac{l_2^2}{m_2}$ to $\frac{l_3^2}{m_3}$. Substituting the length and mass ratios gives $\frac{5^2}{1}$ to $\frac{3^2}{3}$ to $\frac{1^2}{5}$, which simplifies to 25 to 3 to $\frac{1}{5}$, ultimately resulting in 125 to 15 to 1.

The terminal voltage of a source is always less than its electromotive force by an amount equal to the potential drop across its internal resistance, modeled by the relation $V = E - Ir$. Multiplying the internal resistance of 2 ohms by the current of 0.6 amperes yields a voltage drop of 1.2 volts. Subtracting this internal drop from the given emf of 12 volts results in a terminal voltage of 10.8 volts.

The equivalent resistance of a triangular circuit between any two vertices is found by recognizing that two of the resistors are connected in series, and this combined branch is in parallel with the third resistor. Adding the two adjacent resistances gives $4 + 4 = 8$ ohms. Combining this result in parallel with the remaining resistor yields $\frac{8 \times 4}{8 + 4} = \frac{32}{12} = \frac{8}{3}$ ohms.

Symmetry principles allow us to simplify symmetrical resistor networks by recognizing identical current distributions across mirrored branches. The internal section containing node $O$ features resistors $CO$ and $OD$ connected in series, creating a combined path of 2 ohm.

This segment lies in parallel with the branch $CD$ of resistance 1 ohm, yielding an equivalent resistance of $\frac{2}{3}$ ohm. Adding the series resistors $AC$ and $DB$ to this intermediate result produces a total upper path resistance of $\frac{2}{3} + 1 + 1 = \frac{8}{3}$ ohm. Finally, combining this upper network in parallel with the direct branch $AB$ of resistance 2 ohm gives $\frac{(\frac{8}{3}) \times 2}{(\frac{8}{3}) + 2}$, which simplifies to $\frac{16}{14}$ or $\frac{8}{7}$ ohm.

Kirchhoff's first law states that the total electric current entering any junction in a circuit must equal the total current leaving it. This principle relies on the physical reality that electric charge cannot accumulate indefinitely at a single point in a steady-state circuit. Consequently, the net flow of charge into a node must balance out completely, establishing that the rule fundamentally governs the conservation of charge.

The electrical power dissipated in a circuit is expressed by the relation $P = \frac{V^2}{R_{\text{eq}}}$. Three identical resistors connected in series present an equivalent resistance of $3r$, leading to a power dissipation of $\frac{V^2}{3r}$, which establishes that $\frac{V^2}{r}$ equals three times the power. Reconfiguring these same resistors into a parallel network reduces the equivalent resistance to $\frac{r}{3}$. Applying this new equivalent resistance to the power formula yields $P = \frac{V^2}{r/3} = \frac{3V^2}{r}$, which triples the initial power dissipation value to ninety watts.

The electrical resistivity of a material is determined by its macroscopic resistance combined with its physical dimensions, following the relation
$\rho = R \frac{A}{l}$
First, the resistance of the wire is found using Ohm's law by dividing the voltage of 2 V by the current of 4 A, yielding 0.5 \Omega. Substituting this resistance alongside the cross-sectional area of 1 \text{ mm}^2 and the length of 50 \text{ cm} into the resistivity formula results in a resistivity value of 1 × 10^{-6} \Omega \text{ m}.

Electrical resistance depends inversely on the cross-sectional area through the formula $R = \rho \frac{l}{A}$. Since the thicker wire has a resistance of 10 ohms and an area three times that of the thinner wire, the thinner wire must have a resistance three times larger, amounting to 30 ohms. Summing these individual resistances for components connected in series yields a combined total resistance of 40 ohms.

The brightness of bulbs connected in a series circuit depends on the power dissipated, which increases when the total resistance of the circuit decreases under a constant supply voltage. When 40 bulbs are connected in series, the high total resistance restricts the current to a smaller value. Removing one fused bulb reduces the total resistance to that of 39 bulbs, allowing a larger current to flow from the 220 V supply and causing the remaining bulbs to glow more brightly.

Finding the total current in a combination circuit relies on applying Ohm's law alongside the fundamental rules for series and parallel resistors. First, the two resistors with values of 6 ohms connected sequentially along the same path are added directly, giving a combined resistance of 12 ohms. This resulting segment is connected in parallel with the 3 ohm resistor, so the total equivalent resistance is calculated using the formula $R_{eq} = \frac{R_{BC} \times R_A}{R_{BC} + R_A}$ which evaluates to 2.4 ohms. Applying the principal equation for circuit current, $I = \frac{V}{R_{eq}}$, and substituting the source voltage of 4.8 V along with the total resistance of 2.4 ohms gives a final current of 2.0 A.

Finding the equivalent resistance of a circuit requires applying the series formula $R_{s} = R_1 + R_2$ and the parallel formula $R_{p} = \frac{R_a R_b}{R_a + R_b}$. Analyzing the layout reveals that the lowermost resistor on the extreme left and the uppermost resistor on the extreme right are dangling branches that carry no current and remain ineffective. Working inward, the adjacent pairs of vertical resistors are connected sequentially, combining to form two separate branches of 20 ohms each. These two branches lie side-by-side in a parallel arrangement, which simplifies their combined resistance down to 10 ohms. Adding this parallel result to the remaining entry and exit resistors positioned in series along the main line gives a final equivalent resistance of 30 ohms.

The operation of a metre bridge relies on the Wheatstone bridge balancing condition given by $\frac{R}{S} = \frac{l}{100 - l}$. Interchanging the positions of the cell and the galvanometer does not alter the circuit topology or the mathematical ratio required for zero current, meaning the null point remains exactly where it was. Consequently, moving the jockey on either side of this balance point causes the current in the galvanometer to reverse direction, producing both left-sided and right-sided deflections. At the precise balance point, no current flows through the galvanometer at all, yielding zero deflection.

Negative differential resistance occurs in electrical systems where an increase in applied voltage ($V$) paradoxically causes a decrease in current ($I$). Tracing the trajectory of the given graph reveals that along the section $CD$, the current drops as the corresponding voltage continues to rise. This distinct inverse behavior uniquely identifies the region where negative resistance is observed.

The current flowing through the potentiometer wire is determined by dividing the cell electromotive force by the total wire resistance, yielding 0.2 amperes. The potential gradient, or potential difference per unit length, is calculated using the formula $\frac{V}{l} = I \times \left(\frac{R}{l}\right)$. Multiplying the circuit current by the resistance per unit length of 2.5 ohms per meter gives a potential gradient of 0.5 V/m.

Devices that do not maintain a constant resistance and whose current-voltage characteristic curve is non-linear are classified as non-ohmic components. Inside a discharge tube, the conduction of electricity relies on ionized gas particles and cascading collisions rather than a simple metallic conduction lattice. Consequently, the ratio of voltage to current varies with the applied potential, failing to obey Ohm's linear proportionality.

A bridge network of resistors acts as a balanced Wheatstone bridge when the resistances of its opposite arms are equal, which ensures zero current flows through the central branch. Since every individual resistor in the circuit has a value of 2 ohms, the bridge is balanced and the middle resistor can be safely neglected. The remaining circuit reduces to two parallel branches of series resistors, where each branch totals 4 ohms, ultimately yielding an equivalent resistance of 2 ohms between points A and B.

When an electric current flows through a metallic conductor, the free electrons drift in a specific direction under the influence of an electric field with a sluggish average speed known as drift velocity. Despite the rapid propagation of electrical energy, individual charge carriers move quite slowly due to frequent collisions with the lattice ions. Consequently, this drift velocity under normal operating conditions is typically on the order of a fraction of a millimeter per second.

Equivalent resistance of complex circuits is systematically evaluated by simplifying series and parallel branches starting from the farthest boundary and working inward. Along the outer perimeter, the resistors positioned on sides $AF$ and $FE$ are connected end-to-end, resulting in a series combination of $3 + 3 = 6$ ohms. This branch shares nodes with the diagonal resistor $AE$ of $6$ ohms, forming a parallel pair that reduces to $3$ ohms. Extending this successive reduction across the remaining structural segments simplifies the entire network into a final combination connected across terminals $A$ and $B$. Computing this final arrangement yields an equivalent resistance given by $R_{eq} = \frac{(3+3) \times 3}{(3+3)+3}$ which simplifies to $2$ ohms.

This circuit configuration represents a classic Wheatstone bridge, where a central resistor spans diagonally across a symmetric loop connected to a power source. Since the uniform wire has a total resistance of $4$ ohms, bending it into a square distributes an equal resistance of $1$ ohm across each of its four sides, designated as $AB$, $BC$, $CD$, and $DA$. The ratio of resistances in the opposite arms is equal, satisfying the condition for a balanced bridge where no current flows through the $2$ ohm resistor bridging the middle terminals.

Removing this inactive resistor simplifies the network into two parallel paths between the input and output terminals, where the upper and lower branches each present a total resistance of $2$ ohms. The equivalent resistance of these parallel branches is calculated as $R_{\text{eff}} = \frac{2 \times 2}{2 + 2} = 1\ \Omega$. Applying Ohm's law with a battery potential of $2$ volts yields the total current through the formula $I = \frac{2}{1} = 2\text{ A}$.

Finding the total current in a resistor network relies on Ohm's law, given by the formula $I = \frac{V}{R}$. The circuit configuration first places two of the 3-ohm resistors in series, meaning their resistances add directly to give 6 ohms. This combined branch then runs in parallel with the third 3-ohm resistor, creating an overall equivalent resistance of 2 ohms for the entire circuit. Dividing the source voltage of 2 volts by this total resistance of 2 ohms produces a final current of 1 amp.