Current Electricity – NEET UG PhysicsPractice Questions & PYQs

When identical resistors are combined, the equivalent resistance depends entirely on whether the connection is parallel or series. For $n$ resistors of resistance $r$ connected in parallel, the reciprocal of the equivalent resistance is given by $\frac{1}{R} = \frac{n}{r}$, which means the individual resistance can be expressed as $r = nR$. When these exact same resistors are instead connected in series, the total resistance is simply the sum of individual resistances, yielding $R_s = nr$. Substituting the relationship between $r$ and $R$ into the series expression gives $R_s = n(nR) = n^2 R$.

When all four arms of a Wheatstone bridge possess identical resistances, the bridge is perfectly balanced, ensuring that the electric potentials at both ends of the central galvanometer branch are identical. Because no potential difference exists across this central branch, no current flows through it, rendering its resistance inactive in terms of current distribution. Consequently, the circuit simplifies into two parallel branches, each containing two arm resistances in series, and evaluating the parallel combination of these two identical branches using $\frac{2R \times 2R}{2R + 2R}$ results in an equivalent resistance of $R$.

The electrical resistance of a single bulb is given by the relation $R = \frac{V^2}{P}$, yielding $R = 220^2 / 100 = 484$ ohms. When two such identical resistors are placed in series, the combined resistance is the sum, resulting in $R_{\text{eq}} = 484 + 484 = 968$ ohms, and the power drawn from the source is $P_{\text{eq}} = \frac{V^2}{R_{\text{eq}}}$, which evaluates to $220^2 / 968 = 50$ watts. When the bulbs are wired in parallel, the combined resistance drops to half, giving $R_{\text{eq}} = 242$ ohms, and the corresponding power consumption becomes $P_{\text{eq}} = \frac{V^2}{R_{\text{eq}}} = 220^2 / 242 = 200$ watts.

The heat required to boil a fixed quantity of water depends on the applied voltage and the resistance of the heating element according to the relation $H = \frac{V^2 t}{R}$. When two individual coils are used separately, their respective resistances can be expressed in terms of their boiling times as $R_1 = \frac{V^2 t_1}{H}$ and $R_2 = \frac{V^2 t_2}{H}$. Connecting these coils in parallel modifies the equivalent resistance to $\frac{R_1 R_2}{R_1 + R_2}$, which directly translates the combined boiling time through the harmonic sum formula $t_p = \frac{t_1 t_2}{t_1 + t_2}$. Substituting the given times yields $t_p = \frac{10 \times 40}{10 + 40}$, which simplifies to 8 minutes.

A safety fuse functions by interrupting the electrical circuit when current exceeds a safe threshold, which relies on localized Joule heating. To ensure rapid melting and breaking of the circuit before sensitive appliances are damaged, the material must generate sufficient heat quickly and melt at a relatively low temperature. Therefore, a proper fuse wire requires high electrical resistance to produce intense heat combined with a low melting point so that it fuses instantly under excessive current conditions.

The electromotive force of a source represents its open-circuit terminal voltage, meaning the emf is $E = 2.2$ volts. When a load resistance is attached, the current flowing through the closed circuit is governed by $I = \frac{E}{R + r}$, where $r$ is the internal resistance. The terminal potential difference across the external resistor is given by $V = I R$, which equates to $\frac{E R}{R + r} = 1.8$ volts. Substituting the known values gives $\frac{2.2 \times 5}{5 + r} = 1.8$, and solving this linear expression yields an internal resistance of $r = \frac{10}{9}$ ohms.

Electrical resistivity is an intrinsic material property that depends microscopically on free electron collision rates and lattice vibrations. Because these atomic-level interactions intensify as thermal energy rises, the resistivity of a metallic conductor increases with temperature. Dimensions such as length and cross-sectional area alter the total resistance of a specific object, but they have no effect on the inherent specific resistance of the material itself.

The specific resistance of metals decreases with a drop in temperature due to reduced thermal lattice vibrations that facilitate smoother electron flow. In contrast, semiconductors experience an increase in specific resistance as temperature falls because thermal energy becomes insufficient to excite charge carriers into the conduction band. Cooling both materials from 300 K to 60 K therefore lowers the resistivity of the copper conductor while raising the resistivity of the silicon semiconductor.

The potential gradient along a potentiometer wire represents the potential drop per unit length, defined by the relation $\frac{V}{l} = \frac{I \rho}{A}$. Substituting the given values for current, specific resistance, and cross-sectional area gives the expression $(0.1 \times 10^{-7}) / 10^{-6}$. Evaluating this calculation results in a potential gradient of 0.01 V/m.

A balanced Wheatstone bridge has equal resistances in all four arms, meaning the potential difference across the galvanometer branch is zero. Because no current flows through the galvanometer and its series-connected 10 ohm resistor, this middle branch draws no current and does not affect the overall circuit resistance. The remaining network consists of parallel arms that simplify symmetrically, yielding an equivalent battery resistance equal to the individual arm value of 10 ohm.

The electrical power rating of a device operating at a specified voltage is inversely proportional to its resistance, as indicated by the relation $P = \frac{V^2}{R}$. Since both bulbs are designed for the same voltage of 200 V, the bulb with the lower power rating draws less power because it possesses a higher filament resistance. Comparing the 40 W bulb and the 100 W bulb reveals that the resistance of the 40 W bulb is greater than that of the 100 W bulb.

When a secondary battery receives a charging current from an external source, current enters its positive terminal, causing the terminal potential difference to exceed its electromotive force according to the equation $V = E + Ir$. Substituting the given values of electromotive force, charging current, and internal resistance into this relation gives $V = 12 + 60 \times (5 \times 10^{-2})$. Evaluating this expression yields a terminal potential difference of 15 V.

Recognizing the balanced Wheatstone bridge configuration reveals that the central 7 ohm resistor carries no current and can be removed from the analysis. This simplifies the network into two parallel branches connected between the terminals. The first branch combines the 4 ohm and 3 ohm resistors in series for a total of 7 ohms, while the second branch combines the 6 ohm and 8 ohm resistors in series for a total of 14 ohms. Calculating the net resistance of these parallel paths is given by the formula $\frac{1}{R_{eq}} = \frac{1}{4+3} + \frac{1}{6+8}$ which yields $R_{eq} = \frac{14}{3}$ ohms.

Measuring voltage without drawing any current from the circuit ensures that the true open-circuit potential difference is obtained without internal resistance interference. A potentiometer achieves this by balancing the unknown potential against a known adjustable potential drop until a sensitive galvanometer shows zero current flow. Because no current is drawn at the null point, the device effectively acts as an infinite resistance voltmeter during measurement.