Current Electricity – NEET UG PhysicsPractice Questions & PYQs

To achieve maximum power dissipation across an external resistor using multiple identical sources, the batteries must be connected in series to yield a total electromotive force of $2 + 2 = 4$ volts and a combined internal resistance of $1 + 1 = 2$ ohms. Maximum power transfer occurs when the external load resistance matches this total internal resistance, so $R = 2$ ohms. Substituting these values into the power relation yields $P = \left(\frac{4}{2 + 2}\right)^2 \times 2 = 2.0$ watts.

When two opposing voltage sources are connected in a single-loop circuit, the net current is determined by dividing the difference in their electromotive forces by the sum of all internal and external resistances. Calculating this gives the circuit current through the relation
$I = \frac{8-4}{1+2+9} = \frac{4}{12} = \frac{1}{3}\text{ A}$

The potential difference between points $P$ and $Q$ is then found by evaluating the voltage along that specific branch, which is expressed as
$V_{P}-V_{Q}=4-\frac{1}{3} \times 3=3\text{ V}$

The electrical resistance of a device can be determined from its power rating and operating voltage using the power relation
$P = \frac{V^{2}}{R}$
Rearranging this formula to solve for resistance gives R = \frac{V^2}{P}. Substituting the voltage of 220 \text{ V} and power of 60 \text{ W} into the equation yields R = \frac{(220)^2}{60}, which calculates to 807 \Omega.

Ohm's law dictates that the total current supplied by a source equals the system voltage divided by the net equivalent resistance. The process begins by combining the parallel branch resistors, yielding $R' = \frac{4 \times 8}{12} = \frac{8}{3}$ ohms. Summing this result with the remaining series components results in a total equivalent resistance of $R_{eq} = \frac{8}{3} + \frac{1}{3} + 1.5 + 5.5 = 10$ ohms. Applying the relationship $i = \frac{V}{R_{eq}}$ with the battery potential of $5$ volts produces a current of $0.5$ amperes.

Cutting a uniform wire into eight equal segments creates individual pieces each having a resistance of $\frac{R}{8}$. When four of these identical segments are connected in parallel, their combined resistance reduces by a factor of four to become $\frac{R}{32}$. Connecting two such parallel sets in series simply adds their individual resistances together, resulting in a net effective resistance of $\frac{R}{16}$.

Calculating the equivalent resistance of a combined series-parallel network relies on determining the individual parallel blocks before combining them in series. The first parallel section containing the $1\,\Omega$ and $3\,\Omega$ resistors simplifies to $\frac{3 \times 1}{3+1} = \frac{3}{4}$, while the second parallel section with the $2\,\Omega$ and $4\,\Omega$ resistors simplifies to $\frac{2 \times 4}{2+4} = \frac{8}{6}$. Adding these two equivalent resistances in series yields a total network resistance of $\frac{3}{4} + \frac{8}{6} = \frac{25}{12}$.

Applying Ohm's law with the constant source voltage of 50 V across this total resistance gives the main circuit current as $I = \frac{50}{25/12}$, which results in 24 A. This total current splits at the parallel junctions based on the resistance ratios. The branch containing the $1\,\Omega$ resistor carries a current of $\frac{3}{4} \times 24 = 18$, and the branch with the $2\,\Omega$ resistor carries a current of $\frac{4}{6} \times 24 = 16$.

Applying Kirchhoff's current law at junction $C$ allows the branch current through $CD$ to be found by subtracting the current of the second parallel branch from the first parallel branch. Taking the difference between the 18 A and 16 A currents gives a final result of 2.0 A directed from $C$ to $D$.

The macroscopic current in a conductor links directly to the microscopic drift velocity of electrons through the expression $i = n A v_d e$, where the cross-sectional area scales with the square of the wire diameter, making current proportional to $d^2 v_d$. Setting up a ratio between the initial and final states accounts for the change in current from 100 milliamperes to 200 milliamperes and the diameter halving from $d$ to $\frac{d}{2}$. Solving this proportionality yields a new drift velocity of $8 v$.

The electrical power consumed by a heating element connected to a constant voltage source is inversely proportional to its resistance, expressed by the relationship $P = \frac{V^2}{R}$. Since heater A is rated at 1 kW and heater B at 2 kW, the resistance of heater A is twice that of heater B. When connected in series across a fixed voltage, the total power consumed is given by $P_S = \frac{V^2}{R_A + R_B}$, whereas in parallel the total power is $P_P = \frac{V^2(R_A + R_B)}{R_A R_B}$. Evaluating the ratio of power outputs for the series and parallel configurations yields two-ninths.

The electrical resistance of a uniform wire varies with its length according to the relation $R \propto l^2$ when the volume and mass remain constant during stretching. Doubling the length of the original 10 ohm wire increases its resistance by a factor of four, resulting in a total stretched resistance of 40 ohms. Bending this new wire into a circle divides it into two equal semicircular halves of 20 ohms each connected in parallel across any arbitrary diameter. Finding the equivalent resistance of two identical parallel branches yields half of their individual value, bringing the final resistance back to 10 ohms.

A Wheatstone bridge achieves balance when the ratio of resistances across its opposite arms satisfies the fundamental equation $\frac{P}{Q} = \frac{R}{S}$. Applying this principle to the valid circuit configuration, the required balance condition is established through the proportion $\frac{10}{15} = \frac{10}{5 + R_D}$. Since the semiconductor diode actively conducts in this arrangement, it provides a dynamic resistance of 10 ohms, which perfectly satisfies the equation and brings the bridge into equilibrium.

The electrical resistance of a uniform conductor is determined by material resistivity and its dimensions according to the formula $R = \frac{\rho l}{A}$. Dividing a wire into ten equal segments reduces the resistance of each individual piece to one-tenth of the original value. Connecting five of these equal segments in series gives a combined resistance equal to five times the segment resistance, which evaluates to 50 ohms, while connecting the remaining five segments in parallel results in 2 ohms. Adding these two separate series and parallel sub-combinations together in series gives a final total resistance of 52 ohms.

The terminal voltage of a power source is calculated by subtracting the internal potential drop from its total electromotive force, using the relation $V = E - ir$. The current circulating through the circuit is determined by dividing the total electromotive force of 10 volts by the combined total of the 4 ohm external resistor and the 1 ohm internal resistance, resulting in 2 amperes. Multiplying this current by the 1 ohm internal resistance yields an internal potential drop of 2 volts. Subtracting this internal drop from the initial 10 volt electromotive force produces a final terminal voltage of 8 volts.

When a circuit achieves a zero deflection in the galvanometer, it establishes a balanced Wheatstone bridge condition where the ratio of the adjacent resistances equals the ratio of the corresponding wire segments. For a uniform wire of total length 40 cm, equating these resistance ratios gives the relation $\frac{8}{x} = \frac{12}{40-x}$, where $x$ denotes the balancing length from one fixed end. Reducing this expression leads to $80 - 2x = 3x$, which solves to $x = 16$ cm. Because the distance is required from end B, subtracting this length from the total wire length yields $40 - 16 = 24$ cm.

The variation of electrical resistance with temperature is governed by the linear temperature dependence equation $R = R_0(1 + \alpha \Delta T)$, where $\alpha$ represents the temperature coefficient of resistance. Substituting the initial resistance of 2 ohm at $0^{\circ}\text{C}$ and the final resistance of 6.8 ohm at $80^{\circ}\text{C}$ into the formula gives $6.8 = 2(1 + 80\alpha)$. Solving this linear equation for $\alpha$ yields $\frac{6.8}{2} - 1 = 80\alpha$, which simplifies to $3 \times 10^{-2} {}^{\circ}\text{C}^{-1}$.

The displacement current in the space between capacitor plates is mathematically equivalent to the conduction current, defined as the rate of flow of charge $I = \frac{dQ}{dt}$. Since the charge on a capacitor is related to capacitance and potential by $Q = CV$, differentiating both sides with respect to time gives the relation $I = C \left( \frac{dV}{dt} \right)$. Substituting the given values of 2 mA for the displacement current and $4 \, \mu\text{F}$ for the capacitance into this rearranged expression yields $\frac{2 \times 10^{-3}}{4 \times 10^{-6}}$, which simplifies to 500 V/s.

The internal resistance of a cell connected across an external resistance can be determined using the balancing lengths of the potentiometer wire before and after the external load is applied, through the relationship $r = \left( \frac{\ell_o - \ell_c}{\ell_c} \right) R$. Substituting the initial balancing length of 330 cm, the internal resistance of 1 ohm, and the external resistance of 2 ohm into the formula gives an equation where the unknown balancing length is the only variable. Solving this algebraic relation yields a new balancing length of 220 cm.

Equivalent resistance depends heavily on whether resistors are combined in series or parallel, directly altering the current drawn from a constant voltage source according to Ohm's law. When ten identical resistors of resistance $R$ are connected in series, the total resistance is $10R$, resulting in a current of $i = \frac{E}{10R}$. Reconnecting them in parallel reduces the equivalent resistance to $\frac{R}{10}$, which increases the current to $i' = \frac{10E}{R}$. Comparing the new current to the original series current shows that the current increases by a factor of 100.

The electrical resistance of a uniform wire is inversely proportional to the square of its diameter, governed by the relation $R \propto \frac{1}{d^2}$ because cross-sectional area scales with diameter squared. When the diameter of wire B is made three times larger than that of wire A, its cross-sectional area increases by a factor of nine. This nine-fold increase in area causes the resistance of wire B to decrease to one-ninth of wire A's resistance. Dividing the original resistance of 81 ohms by nine gives a final resistance of 9 ohms for wire B.

Applying Kirchhoff's voltage law, the current in a single-loop circuit is calculated by dividing the net electromotive force by the total resistance, represented by the formula $i = \frac{10-5}{10}$. Simplifying this fraction results \in $i = \frac{5}{10}$, which yields a magnitude of 0.5 A. Because the higher potential dominates the loop, the charge flows \in a clockwise direction, routing from $A$ to $B$ through $E$.

The resistance value of a carbon resistor expressed in standard form identifies each digit and power of ten with a specific color band according to the standard resistor color code chart. Rewriting the given resistance value as $22 \times 10^3 \, \Omega$ reveals that the first two bands represent the digits two and two, while the power of ten multiplier is $10^3$. In the standard color coding scheme, the multiplier $10^3$ corresponds directly to the color orange.