Current Electricity – NEET UG PhysicsPractice Questions & PYQs

The equivalent resistance of identical resistors achieves its maximum value in a series configuration and its minimum value in a parallel configuration. For $n$ resistors of individual resistance $R$, the series resistance is $R_s = n R$, while the parallel resistance is $R_p = \frac{R}{n}$. Taking the ratio of the maximum resistance to the minimum resistance yields $\frac{n R}{R/n}$, which simplifies to $n^2$.

Good electrical conductors rely on a crystal lattice of positive ions surrounded by a mobile sea of delocalized valence electrons. These free-moving charge carriers can drift easily across the entire atomic structure under the influence of an electric field. This unique electron-sea architecture accounts for both the high electrical and thermal conductivity characteristic of metals.

Good electrical conductors are substances that permit the free movement of valence electrons, allowing electric charge to flow through them with minimal resistance. Metals naturally exhibit high conductivity due to their lattice structure and delocalized electrons, making copper, silver, and gold premier metallic conductors. Other options contain semiconductors like germanium or silicon, or ionic compounds and insulators like sodium chloride and diamond, which lack this uniform high metallic conductivity.

When a charged capacitor discharges through a resistor, the total thermal energy released as heat is equal to the initial electrostatic potential energy stored in the capacitor, as expressed by the equation
$E = \frac{1}{2} C V^{2}$
Substituting the capacitance of 4 \mu\text{F} and voltage of 400 \text{ V} into this relation gives E = \frac{1}{2} × 4 × 10^{-6} × (400)^2, which evaluates to 0.32 \text{ J} of heat produced across the 2 \text{ k}\Omega resistance.

Joule's law of heating relates the heat produced in a conductor to the current, resistance, and time through the relation $H = I^2 R t$. Rearranging this expression to isolate resistance gives $R = \frac{H}{I^2 t}$. Substituting the given heat of 80 J, current of 2 A, and time of 10 seconds into this formula yields $R = \frac{80}{2^2 × 10}$, which evaluates to 2 ohms.

A meter bridge operates on the balanced Wheatstone bridge principle where the ratio of resistances in the two gaps equals the ratio of their corresponding balancing lengths on the wire. Given a balancing length of 20 cm from the left end, the remaining length on the right side is 80 cm. Applying the bridge formula $\frac{P}{Q} = \frac{l}{100 - l}$ with the known standard resistance of 1 ohm in the right gap yields an unknown resistance of 0.25 ohms.

The maximum power transfer theorem dictates that a power source delivers maximum power to an external load when the resistance of the load matches the internal resistance of the source. Since the internal resistance of the battery is given as 0.5 ohms, the variable resistance must be set to the exact same value. Therefore, the required external resistance is 0.5 ohms to satisfy this condition.

Power dissipation in a resistor is dictated by the formula $P = I^2 R$, where current remains constant throughout any series connection. Because the power is directly proportional to resistance under a common current, the ratio of power dissipated mirrors the ratio of the resistances themselves. Given that the individual resistances share a proportion of one to two, the resulting power consumption across them maintains that exact same proportion.

To safely handle a total current of 4 amperes using components with a 1 ampere individual rating, the network must consist of four parallel branches, each carrying 1 ampere of current. Achieving an equivalent resistance of 5 ohms for this parallel combination means each individual branch must have a resistance calculated as $5 \times 4 = 20$ ohms. Since each available resistor has a value of 10 ohms, exactly two resistors must be connected in series within each of the four branches to reach 20 ohms, resulting in a total requirement of eight resistors.

Converting a galvanometer into a wider-range ammeter requires a low resistance shunt connected in parallel, such that the potential drop across both branches is equal, expressed as $I_S S = I_G G$. The current diverted through the shunt is effectively the full ammeter range minus the tiny galvanometer current, which is nearly 10 amperes. Using the parallel condition with a galvanometer resistance of 100 ohms and a full-scale current of 0.001 amperes yields a required shunt resistance of 0.01 ohms.

Ohm's law, $V = I R$, establishes the core mathematical relationship required to determine the unknown potential drop in the circuit. The voltage across the 3 ohm resistor carrying 0.8 A is found to be 2.4 V. Since the 6 ohm resistor is connected in parallel with this branch, it shares this 2.4 V potential difference, drawing a current of 0.4 A. Combining the individual branch currents gives a total circuit current of 1.2 A. Multiplying this total current by the resistance of the 4 ohm resistor yields the final potential drop of 4.8 V.

Electric current is defined as the total charge passing through a cross-section per unit time, quantified by the equation
$I = \frac{q}{t} = \frac{n e}{t}$
Given an electron flow rate of 10^7 electrons per second and the elementary charge of 1.6 × 10^{-19} \text{ C}, multiplying these quantities together yields a current of 1.6 × 10^{-12} \text{ A}.

The resistance of an electrical appliance is determined by its rated voltage and power via $R = \frac{V^2}{P}$, and power consumed at a different operating voltage is given by $P = \frac{V_{\text{s}}^2}{R}$. Using the bulb's rated specifications, its fixed internal resistance is calculated as $\frac{200^2}{100}$. When this bulb is connected to a lower supply voltage, the actual power consumption is found by dividing the squared supply voltage by this resistance, yielding sixty-four watts.

Electrical resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area, expressed through the formula $R = \rho \frac{l}{A}$. For the first wire, the resistance scales directly with $\frac{l}{A}$. The second wire increases both length and inverse area, resulting in a resistance scaling of $\frac{2l}{\frac{A}{2}} = 4\frac{l}{A}$. Meanwhile, the third wire reduces the length and doubles the area, yielding a resistance scaling of $\frac{\frac{l}{2}}{2A} = \frac{1}{4}\frac{l}{A}$. Comparing these proportionality factors demonstrates that the third wire achieves the smallest resistance value.

The dependence of resistance on temperature is expressed through the relation $R = R_0(1 \pm \alpha \Delta T)$, and according to Ohm's law, current is inversely proportional to equivalent resistance. In circuit A, where the resistors are connected in series, the total resistance is calculated by adding the individual resistances, yielding $R_0(1+\alpha \Delta T) + R_0(1-\alpha \Delta T) = 2R_0$. Because the temperature-dependent terms completely cancel out, the total resistance of circuit A remains constant, which means the current $I_A$ also remains constant as temperature increases.

For circuit B, where the resistors are connected in parallel, the equivalent resistance is evaluated as $\frac{R_0(1-\alpha^2 (\Delta T)^2)}{2}$. As the temperature rises, the subtraction of the growing squared temperature term causes this equivalent resistance to diminish. Since a reduced equivalent resistance draws a larger flow under a constant voltage, the current $I_B$ increases with temperature.

The total resistance of the circuit is found by adding the load resistance and the internal resistance, yielding 4 ohms. The current flowing through the circuit is determined using Ohm's law via $I = \frac{\varepsilon}{R + r}$, which gives 0.5 amperes. The terminal voltage across the cell is then calculated using the relation $V = \varepsilon - Ir$, resulting in a voltage of 1.95 V.

Power delivered by a fixed voltage source is inversely proportional to resistance through the expression $P = \frac{V^2}{R}$. Cutting the heating coil in half halves its individual resistance, and connecting these two halves in parallel cuts the equivalent resistance to a quarter of the original value. Because power scales inversely with resistance, reducing the total resistance fourfold increases the energy liberated per second to four times the initial rating, resulting in 400 watts.

Power dissipation in a resistive device depends on the square of the operating voltage because resistance is kept constant, represented through the formula $P = \frac{V^2}{R}$. Determining the constant resistance from the rated values gives $R = \frac{V_1^2}{P_1}$, which evaluates to 121 ohms. Substituting this resistance and the reduced voltage into the power expression yields $P_2 = \frac{V_2^2}{R}$, calculating out to approximately 331 watts.

The heat generated in a conductor by an electric current follows the relation $H = I^2Rt$, while the resulting temperature change is determined by $H = ms\Delta T$. Equating these expressions reveals that the temperature rise is proportional to the square of the current, represented as $\Delta T \propto I^2$. When the current is doubled, the squared dependence causes the temperature increase to scale by a factor of four. Multiplying the initial five degree temperature rise by this factor gives a final temperature change of twenty degrees Celsius.

Electrical energy can be expressed by multiplying power by time through the relation $E = P \times t$. Substituting one thousand watts for the power and three thousand six hundred seconds for one hour yields $1000 \times 3600$ joules. Simplifying this product results in $36 \times 10^{5}$ joules.