Current Electricity – NEET UG PhysicsPractice Questions & PYQs

The microscopic drift velocity of charge carriers inside a current-carrying conductor is determined by the fundamental current equation $I = n e A V_d$ which links macroscopic current to electron density, elementary charge, and cross-sectional area. Rearranging this expression to solve for drift velocity yields $V_d = \frac{I}{n e A}$. Substituting the given values for current, electron concentration, electronic charge, and the circular cross-sectional area calculated from a radius of 1 millimeter leads to a final drift velocity expression of $6.25 × 10^3 / \pi$ meters per second.

Capacitors connected in a parallel arrangement combine through the direct summation formula $C_{eq} = C_1 + C_2$. Inspection of the circuit diagram reveals that both components share direct connections across the same two nodes, confirming the parallel setup. Substituting the individual capacitances of 5 microfarads and 15 microfarads into the expression yields $C_{eq} = 5 + 15$, resulting in a total equivalent capacitance of 20 microfarads.

Infinite resistor networks rely on the principle of self-similarity, meaning that removing the very first repeating section leaves the remaining infinite ladder with the exact same total equivalent resistance. Denoting this total resistance as $x$, the circuit can be analyzed by combining a parallel branch of $x$ and a 1 ohm resistor in series with a 2 ohm resistor, which together equate back to $x$. Expressing this relationship gives the equation $\frac{x \times 1}{x+1}+2=x$. Simplifying this expression leads to the quadratic equation $x^2-2x-2=0$. Solving for $x$ yields roots of $1 \pm \sqrt{3}$, and discarding the physically impossible negative value leaves a final equivalent resistance of $1+\sqrt{3}$ ohms.

Electrical resistance in metallic conductors arises from collisions between free electrons and vibrating lattice ions, which become more frequent as thermal energy rises. Consequently, raising the temperature amplifies these atomic vibrations and increases the resistance of conductors. Conversely, semiconductors possess a limited number of thermally available charge carriers, and supplying heat excites electrons across the forbidden band to generate more charge carriers, leading to a net decrease in their resistance.

The thermal energy developed in a resistor is given by the relation $H = \frac{V^2}{R} t$, where $V$ is the potential difference, $R$ is the resistance, and $t$ is time. Since the resistors are connected in parallel, they share an identical potential difference across their terminals for any given time duration, making the thermal energy dissipated inversely proportional to the resistance such that $\frac{H_1}{H_2} = \frac{R_2}{R_1}$. Substituting the resistance values of $200$ ohms and $100$ ohms directly into this relationship yields a thermal energy ratio of $2$ to $1$.

Analyzing a mixed circuit requires breaking it down into series and parallel combinations to find the total equivalent resistance and current drawn from the source. The sliding contact divides the total wire resistance into a segment of resistance $\frac{R_0}{4}$ connected in parallel with the external resistor $R$, while the remaining wire segment of resistance $\frac{3 R_0}{4}$ acts in series with this parallel combination across the voltage source $V_0$.

First, determine the equivalent resistance of the parallel branch formed by the resistor $R$ and the first section of the wire, which evaluates to $\frac{(\frac{R_0}{4}) R}{\frac{R_0}{4} + R}$. Adding the resistance of the second wire segment to this parallel equivalent yields the total circuit resistance $R_{eq} = \frac{(\frac{R_0}{4}) R}{\frac{R_0}{4} + R} + \frac{3 R_0}{4}$, simplifying to $\frac{R_0(16 R + 3 R_0)}{4(R_0 + 4 R)}$. Dividing the source voltage $V_0$ by this total equivalent resistance gives the main circuit current $I = \frac{V_0}{R_{eq}}$. Finally, multiplying this current by the parallel branch resistance gives the potential drop across the resistor $R$, resulting in the expression $\frac{4 V_0 R}{3 R_0 + 16 R}$.

The precision of a Wheatstone bridge relies on its electrical sensitivity, which reaches its maximum when the bridge arms have comparable magnitudes. For the most accurate determination of an unknown resistance, the ratio arms $P$ and $Q$ should be chosen to be approximately equal to each other. Furthermore, keeping these resistances small ensures maximum sensitivity to slight imbalances, ultimately yielding the most precise measurement possible.

The current flowing through a circuit powered by a single cell is determined by the total resistance, calculated as $I = \frac{E}{R + r}$, where $E$ is the emf, $R$ is the external resistance, and $r$ is the internal resistance. The terminal potential difference across the cell drops below its emf due to the internal voltage drop, following the relation $V = E - Ir$. Substituting the evaluated current and internal resistance values into this equation accounts for the potential lost across the internal source, resulting in a terminal potential difference of $3.75$ volts.

Electrical conductance quantifies how easily an electric current flows through a material and is defined mathematically as the inverse of electrical resistance. While resistance impedes the movement of charge carriers, conductance directly measures the material's ability to facilitate this flow. Thus, a higher resistance naturally corresponds to a lower conductance value, following a reciprocal relationship.

Current density is related to the electric field through conductivity as $j = \sigma E$, where conductivity can be expressed in terms of geometric parameters and resistance as $\sigma = \frac{L}{R A}$. Combining these expressions yields $j = \frac{L E}{R A}$, where $L$ is length, $E$ is the electric field, $R$ is resistance, and $A$ is the cross-sectional area. Substituting the area $\pi r^2$ alongside the given values into the formula gives $j = \frac{10 \times 10}{10 \times \pi \left(\frac{10^{-2}}{\sqrt{\pi}}\right)^2}$, which simplifies to a current density of $10^5$ amperes per square meter.

The balancing length of a potentiometer wire varies directly with the electromotive force of the connected cell according to the relation $\frac{E_1}{E_2} = \frac{l_1}{l_2}$. Substituting the initial electromotive force of 1.5 V, the replacement electromotive force of 2.5 V, and the initial balancing length of 36 cm into this proportion gives a numerical ratio of 1.5 divided by 2.5 equal to 36 divided by the unknown length. Solving this algebraic relationship determines that the new balance point occurs at a length of 60 cm.

The equivalent resistance of identical resistors combined in parallel is given by the formula $R_p = \frac{R}{4}$, which allows us to find the individual resistance of each wire. Substituting the given parallel resistance of 0.25 ohms into this relation establishes that each wire has a resistance of 1 ohm. Reconfiguring these same four identical resistors into a series circuit increases the total resistance through the formula $R_s = 4R$, resulting in an effective resistance of 4 ohms.

The microscopic behavior of current in a metallic conductor is governed by fundamental relations connecting charge carrier dynamics and electric fields. Drift velocity represents the average directional speed acquired by electrons under an electric field, expressed as $v_{d} = \frac{e E \tau}{m}$, which links it to charge, field strength, relaxation time, and mass. Electrical resistivity is defined as the ratio of the electric field to the current density, written as $\rho = \frac{E}{J}$. Rearranging microscopic conductivity expressions allows the relaxation period to be expressed as $\tau = \frac{m}{n e^{2} \rho}$. Finally, current density measures the electric current passing per unit cross-sectional area and is formulated \in terms of charge carrier concentration, electronic charge, and drift velocity as $J = n e v_{d}$.

When current encounters a parallel junction formed by resistances $r_{2}$ and $r_{3}$, the incoming total current $i_{1}$ divides such that the voltage drop across both parallel branches stays identical. Equating these potential drops yields the relation $i_{2}r_{2} = i_{3}r_{3}$, which allows us to express $i_{2}$ as $\frac{i_{3}r_{3}}{r_{2}}$. According to the junction rule, the total current equals the \sum of the individual branch currents, leading to $i_{1} = i_{2} + i_{3}$. Substituting the expression for $i_{2}$ into this \sum gives $i_{1} = i_{3}\left(\frac{r_{3}}{r_{2}} + 1\right)$, and rearranging the terms smoothly isolates the desired ratio as $\frac{i_{3}}{i_{1}} = \frac{r_{2}}{r_{2}+r_{3}}$.

The meter bridge balance condition equates the ratio of resistances to the ratio of balancing lengths through the formula $\frac{P}{Q} = \frac{l_1}{l_2}$. Applying this principle with the 10 ohms resistance in the right gap and the 3 to 2 length ratio determines the unknown wire resistance $P$ as 15 ohms. Equating the ratio of this wire's resistance to a 1 ohm segment with their respective lengths gives a second segment length of 0.1 m, which corresponds to 1.0 × 10^-1 m.

Resistor color coding relies on a standardized band system where the first two bands indicate significant digits, the third band acts as a multiplier, and the fourth band specifies the tolerance. Following this convention, the yellow band corresponds to the digit four and the violet band corresponds to the digit seven, establishing the base numerical value. The third band of brown provides a multiplier of ten to the first power, while the gold band indicates a tolerance of five percent. Combining these components into the mathematical expression $R = 47 \times 10^1 \pm 5\%$ results in a final resistance value of 470 ohms with a tolerance of 5 percent.

The temperature coefficient of resistance indicates how a material's electrical resistance responds to temperature changes, where a negative coefficient means resistance decreases as temperature rises. This behavior is characteristic of insulators and semiconductors because thermal energy excites more charge carriers into the conduction band, enhancing overall conductivity. Conversely, metals exhibit a positive temperature coefficient because increased atomic lattice vibrations at higher temperatures hinder electron flow.

Carrier mobility represents the drift velocity acquired per unit electric field and is defined by the formula $\mu = \frac{v_d}{E}$. Substituting the given drift velocity of 7.5 × 10^-4 m s^-1 and the electric field of 3 × 10^-10 V m^-1 into this expression gives a fraction of those two values. Performing this division yields a mobility value of 2.5 × 10^6 m^2 V^-1 s^-1.

The resistivity of a metal is related to temperature by the expression $\rho = \rho_0 [1 + \alpha (T - T_0)]$, which maintains a nearly linear dependence at ordinary temperatures. However, at temperatures much lower than 0°C, the thermal agitation of the crystal lattice drops drastically, causing this linear model to fail. Under these conditions, the graph deviates considerably from a straight line and bends sharply as it approaches absolute zero.

The operation of a metre bridge relies on the Wheatstone bridge principle, where the electrical network achieves balance when the potential difference across the detector becomes zero. Because of the inherent symmetry of a balanced bridge network, swapping the connections of the power source and the galvanometer does not disturb this equilibrium. As a result, the bridge continues to operate normally, preserving the original balance condition of $\frac{P}{Q} = \frac{l_1}{l_2}$.