Current Electricity – NEET UG PhysicsPractice Questions & PYQs

The time taken by an electric appliance to heat a substance is determined by equating the electrical power input to the rate of thermal energy absorption, governed by the equation $VIt = m s \Delta T$. The electrical power is calculated as 220 V multiplied by 4 A, yielding 880 W, while the heat required to raise 1 kg of water through a temperature change of 80 degrees Celsius is calculated as $1000 \times 1 \times 80 \times 4.2 \ \mathrm{J}$, totaling 336,000 J. Dividing this total thermal energy requirement by the power of 880 W gives approximately 382 seconds, which converts to 6.3 minutes.

The internal resistance of a cell connected to a potentiometer is related to the balancing lengths obtained in an open circuit and a closed circuit through the formula $r = \left(\frac{l_1 - l_2}{l_2}\right) R$. Substituting the open-circuit balancing length of 110 cm and the closed-circuit balancing length of 100 cm alongside the external resistance of 10 ohms gives $r = \left(\frac{110 - 100}{100}\right) \times 10$. Evaluating this expression simplifies the fraction to $\frac{10}{100}$, which multiplies by 10 to yield an internal resistance of 1.0 ohm.

Electrical resistance is determined by the formula $R = \rho \frac{l}{A}$, where stretching a wire increases its length while proportionally decreasing its cross-sectional area to keep the total volume constant. When the length is increased by ten percent, the new length becomes $1.1 l$ and the corresponding cross-sectional area reduces to $\frac{A}{1.1}$. Substituting these scaled dimensions into the resistance formula results in $R' = \rho \frac{1.1 l}{A / 1.1}$, which expands to $1.21$ times the initial resistance while the specific resistance remains entirely unchanged.

Parallel resistors share an identical potential drop, meaning the current through the 3 ohm resistor can be determined using the equation $4 \times 1 = 3 \times I_1$ which yields $I_1 = \frac{4}{3}$ amperes when the 4 ohm branch carries 1 amp. Combining these parallel paths gives an equivalent resistance of 12/7 ohms, resulting \in a total current of 7/3 amperes and a potential difference of 4 volts between points P and M. Following the circuit path into the arm connected through N, the total current flowing through this section evaluates to the equation $I' = \frac{16}{5}$ amperes. Multiplying this current by the 1 ohm resistance in that final segment produces the potential difference between points M and N as 3.2 volts.

Parallel branches in a circuit always share an identical potential difference, which makes it straightforward to determine branch currents and calculate power using the relation $P = I^2 R$. Since the $2$ ohm, $4$ ohm, and the combined $1$ ohm and $5$ ohm resistors are connected \in parallel, the potential difference across the entire network equals the voltage drop across the first branch, which is calculated as $3$ amp multiplied by $2$ ohms, giving $6$ volts. This same potential difference drives current through the third branch, where the total resistance is $1$ ohm plus $5$ ohms, resulting \in a branch current of $1$ amp. Multiplying the square of this current by the resistance of the $5$ ohm resistor yields a dissipated power of $(1)^2 × 5$, which equals $5$ watts.

To find the total power dissipated, we first determine the equivalent resistance of the parallel combination of the $6 \Omega$ and $3 \Omega$ resistors using the formula $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$, which evaluates to an equivalent resistance of $2 \Omega$.

Combining this result with the $4 \Omega$ series component establishes the total current flowing through the system via Ohm's law as $I = \frac{18}{2 + 4} = 3 \text{ A}$. Multiplying this total current by the source voltage using the power equation $P = V \times I$ yields the final dissipated power of $54$ watts.

A Wheatstone bridge is balanced when the ratio of the resistances in the ratio arms equals the ratio of the standard resistance to the equivalent unknown arm resistance, expressed as $\frac{P}{Q} = \frac{R}{X}$. Substituting the given arm values of 2 ohms into the balance condition reveals that the equivalent parallel resistance $X$ must equal 2 ohms. Because this equivalent resistance is formed by connecting a 6 ohm resistor in parallel with the unknown resistance $S$, the relationship is given by $\frac{1}{2} = \frac{1}{S} + \frac{1}{6}$, which solves to an unknown resistance value of 3 ohms.

The principle of conservation of electric charge dictates that the net current entering any junction must equal zero, forming the foundation of Kirchhoff's first law. Conversely, Kirchhoff's second law relies on the conservation of energy, ensuring that the total potential change around any closed electrical loop sums to zero. Together, these fundamental conservation laws govern circuit analysis by maintaining steady-state charge and energy distribution throughout complex networks.

Power dissipation in any resistor relies on the relation $P = i^2 R$, where current and resistance dictate the thermal energy released.

Within the circuit, the $3\text{ ohm}$ and $1\text{ ohm}$ resistors are connected in series, creating a branch resistance of $4\text{ ohm}$ that runs in parallel with the $8\text{ ohm}$ resistor. Because parallel branches divide current inversely relative to their resistance, the current flowing through the $4\text{ ohm}$ branch is twice that of the $8\text{ ohm}$ branch.

Relating their power outputs through the equation $P_1 / P_2 = \frac{i_1^2 \times 3}{i_2^2 \times 8}$, and substituting the current expressions $i_1 = 2i/3$ and $i_2 = i/3$, the ratio simplifies to $3/2$. Multiplying this fraction by the known $2\text{ watt}$ dissipation of the $8\text{ ohm}$ resistor yields a final power dissipation of $3\text{ watts}$ across the $3\text{ ohm}$ resistor.

To find the behavior of the conducting wire connected between points $A$ and $B$, we begin by calculating the electric potentials at $A$ and $B$ with respect to point $C$. The current through the corresponding arm is given by $I = \frac{V}{8}$, which establishes a potential difference of $V_C - V_A = \frac{V}{2}$.

For the other section of the circuit, the current is $I' = \frac{V}{4}$, leading to a potential difference of $V_C - V_B = \frac{V}{4}$. Combining these results gives the potential difference between the two points as:
$V_A - V_B = \frac{V}{4} - \frac{V}{2} = -\frac{V}{4}$
Since $V_A - V_B < 0$, the potential at $B$ exceeds that of $A$, dictating the direction of current flow through the newly introduced wire.

The current flowing through a series circuit containing two cells of identical electromotive force and an external resistance $R$ is expressed by the formula $i = \frac{2E}{r_{1}+r_{2}+R}$. The terminal potential difference across the first cell is given by $E - ir_{1}$, which is stated to be zero, meaning the circuit current can alternatively be written as $i = \frac{E}{r_{1}}$. Equating these two expressions yields $\frac{2E}{r_{1}+r_{2}+R} = \frac{E}{r_{1}}$, which simplifies directly to $2r_{1} = r_{1} + r_{2} + R$. Rearranging this equation isolates the external resistance as $R = r_{1} - r_{2}$.

Bending a uniform wire of resistance $R$ into a complete circle divides its total resistance equally into two semicircular halves, each possessing a resistance of $\frac{R}{2}$. When looking across diametrically opposite points, these two identical halves form a parallel combination in the circuit. Applying the parallel resistance formula for two equal branches yields $\frac{1}{R_{\text{eq}}} = \frac{2}{R} + \frac{2}{R}$, which simplifies to an equivalent resistance of $\frac{R}{4}$.

Finding the total current drawn from the voltage source requires determining the equivalent resistance of the circuit using series and parallel combinations. The circuit configuration consists of two parallel branches connected across the voltage V. The first branch contains a 4 ohm resistor in series with a 2 ohm resistor, combining to give a total branch resistance of 6 ohms. The second branch contains a 6 ohm resistor in series with a 3 ohm resistor, yielding a resistance of 9 ohms for that path.

Combining these two parallel branches gives the equivalent resistance of the network, resulting in $R_{eq} = \frac{18}{5}$ ohms. Applying Ohm's law to find the main circuit current using the relation $i = \frac{V}{R_{eq}}$ leads directly to the final expression $i = \frac{5V}{18}$.

Electrical power dissipated in a resistor is related to current and resistance through the relation $P = i^2 R$. Substituting the given maximum power of 1 watt and current of 5 amperes into this equation gives $1 = (5)^2 R$. Solving for resistance yields $R = \frac{1}{25}$, which evaluates to 0.04 ohm.

When two unequal sources are connected in opposition, the net electromotive force driving the loop is determined by Kirchhoff's voltage law as $18 - 12 = 6$ volts. The total resistance opposing this current is the \sum of the individual internal resistances, calculated as $2 + 1 = 3$ ohms. Applying Ohm's law, the current flowing through the circuit is $I = \frac{6}{3} = 2$ amperes. The potential difference across the first battery decreases due to its internal resistance, representing a voltage drop of $2 × 2 = 4$ volts. Subtracting this internal drop from the total electromotive force of that battery gives a final voltmeter reading of $18 - 4 = 14$ volts.

The electrical resistance of an appliance designed for a specific power rating depends directly on its operating voltage through the relation $P = \frac{V^2}{R}$. Since both bulbs share the same 60 watt power rating, equating the power expressions for the Indian and American specifications allows us to relate their resistances. Specifically, the ratio of the voltages squared determines the ratio of their resistances, expressed mathematically as $\frac{220^2}{R} = \frac{110^2}{R_U}$. Solving this equation shows that the resistance required for the American market is one-fourth of the resistance used in India, yielding $\frac{R}{4}$.

This circuit functions as a balanced Wheatstone bridge, meaning the opposing junction nodes sit at identical potentials and no current passes through the bridging resistor. Removing this inactive resistor simplifies the network into two parallel paths connected across the $V$ volt battery, where each path offers a combined resistance of $2R$. Consequently, the total equivalent resistance between terminals $A$ and $B$ evaluates to $R$, resulting in a total source current of $I = \frac{V}{R}$. Because the upper path $AFCEB$ and the lower path share identical resistance values, the total current splits evenly between them, leaving the current flowing through $AFCEB$ at precisely $\frac{V}{2R}$.

The power consumed by a resistive device operating at a specific voltage is inversely related to its resistance, which can be determined from its rated voltage and power via $R = \frac{V^2}{P}$. When three identical bulbs are connected in series, their combined resistance is three times the resistance of a single bulb. Connecting this series combination across the 200 volt supply results in a total power consumption calculated using $P = \frac{V^2}{R_{\text{eq}}}$. Substituting the series resistance expression into this power formula simplifies the calculation to 20 watts.

Electrical resistance depends directly on length and inversely on the cross-sectional area as expressed by $R = \rho \frac{l}{\pi r^2}$. When both the length and radius are doubled, the new resistance becomes proportional to $\frac{2l}{\pi(2r)^2}$, which simplifies to half of the original value. Meanwhile, the specific resistance or resistivity is an intrinsic material property that depends solely on the substance composition and temperature rather than its physical dimensions, remaining completely unchanged.

The potential difference across a segment of a uniform wire depends on the current flowing through it and the resistance of that specific portion. Since the resistance of a wire is proportional to its length, the resistance of the 50 cm segment is calculated using the ratio of lengths, yielding $R_1 = 100 \times \frac{0.5}{3}$ ohms. The total current circulating in the circuit is determined by the total voltage and total resistance via $I = \frac{V}{R_{\text{total}}}$, which evaluates to $\frac{6}{100}$ amperes. Multiplying this current by the segment resistance gives the potential difference, computed as $V_{\text{seg}} = I R_1$ resulting in 1 volt.