Current Electricity – NEET UG PhysicsPractice Questions & PYQs

The total power consumed by a circuit connected to a constant electromotive force is inversely proportional to its equivalent resistance, determined by the formula $P = \frac{E^2}{R_{eq}}$. When all six identical bulbs are fully active, the equivalent resistance of the network combines to $R_{eq} = \frac{2R}{3}$, leading to an initial power consumption of $P_1 = \frac{3E^2}{2R}$. Altering the active components so that only two bulbs from section A and one bulb from section B remain glowing adjusts the equivalent resistance to $R_{eq} = \frac{3R}{2}$, which results \in a modified power consumption of $P_f = \frac{2E^2}{3R}$. Comparing the initial power to this subsequent power through a direct ratio yields the final value of $9 : 4$.

Ideal voltmeters possess infinite internal resistance to prevent drawing any current, whereas ideal ammeters feature zero resistance so they do not impede the flow of charge. Because of these distinct characteristics, incorporating them into the circuits does not alter the fundamental resistance path of the system. Applying Ohm's law in the form of $i = \frac{V}{R}$, the current flowing through both configurations evaluates to $1\text{ A}$, meaning $i_1 = i_2$. Furthermore, the potential difference measured across the resistor in each arrangement is determined by the equation $V = iR$, which yields $10\text{ volts}$ for both meters, establishing that $V_1 = V_2$.

Excessive electrical current generates dangerous levels of heat that can damage sensitive appliances or cause electrical fires. A specialized sacrificial component is typically integrated into the circuit, designed with a very low melting point and high resistance compared to the surrounding wiring. When current exceeds safe operating thresholds, this specific wire rapidly heats up and melts, breaking the circuit and stopping the flow of electricity before further damage occurs.

When identical cells are combined in series, the total electromotive force scales as $n\varepsilon$ while the total internal resistance increases proportionally to $nr$. Applying Ohm's law to the short-circuited terminals gives the current expression $I = \frac{n\varepsilon}{nr}$, which algebraically simplifies to $I = \frac{\varepsilon}{r}$ as the variable $n$ cancels out entirely. Because the resulting current is completely independent of the number of cells, the correct graphical representation is a horizontal straight line showing a constant value.

Ohm's law relates the total circuit current to the electromotive force and the equivalent resistance of the network. For the initial series configuration, the total resistance is $nR + R$, leading to the current expression $I = \frac{E}{nR + R}$. When the resistors are rearranged in parallel, the equivalent resistance becomes $\frac{R}{n}$, resulting in a new current of $10I = \frac{E}{\frac{R}{n} + R}$. Dividing the parallel current equation by the series current equation eliminates the emf and internal resistance, leaving the algebraic ratio $10 = \frac{(n + 1)R}{(\frac{1}{n} + 1)R}$, which simplifies directly to $n = 10$.

The standard resistor color-coding system translates the numerical value and tolerance of a component into colored bands based on specific digit assignments. Converting the given resistance yields forty-seven followed by three zeros with a ten percent tolerance band. Matching these figures to the standard chart where four represents yellow, seven represents violet, three represents the orange multiplier, and ten percent corresponds to silver gives the required sequence.

Electrical resistance depends directly on length and inversely on cross-sectional area, expressed through the formula $R = \frac{\rho l}{A}$. When a wire is reshaped by melting and stretching, its total volume remains strictly constant while its length increases by a factor of $n$. Expressing the area in terms of volume and length transforms the resistance equation into $R = \frac{\rho l^2}{\text{volume}}$, which demonstrates that resistance varies directly with the square of the length. Multiplying the original resistance by the square of the stretching factor yields the new resistance value of $n^2R$.

Accurate electrical measurements require eliminating internal resistance errors that standard voltmeters introduce by drawing a small operating current from the source. A potentiometer achieves this precision by balancing the unknown potential against a known potential gradient until the galvanometer registers zero deflection. Operating under this null condition ensures that absolutely no current is drawn from the cell being tested, allowing for a true measurement of its electromotive force.

The balancing length in a potentiometer is directly proportional to the electromotive force being balanced, represented by the relation $E = Kl$, where $K$ is the potential gradient of the wire. When the two cells are connected to support one another, the total emf is the sum of their individual emfs, which balances at a length of 50 cm, establishing the expression $E_1 + E_2 = K(50)$. Reversing one cell opposes their effects, making the net emf equal to their difference, which balances at a shorter length of 10 cm to give $E_1 - E_2 = K(10)$. Taking the ratio of these two equations eliminates the potential gradient and simplifies to 5 to 1 for the sum versus the difference, which when solved yields the final ratio of 3 to 2 for the emfs of the two cells.

Applying Kirchhoff's voltage law along the path from point A to point B allows us to track the changes in electrical potential. As we move in the direction of the current of 2 A, we account for the potential drops across each circuit element, which include the 2 ohm resistor, the 3 V source, and the 1 ohm resistor. Expressing this principle yields the relation $V_A - I(2) - 3 - I(1) - V_B = 0$. Substituting the given current of 2 A into this equation gives $V_A - 2 \times 2 - 3 - 2 \times 1 - V_B = 0$, which simplifies to $V_A - 9 - V_B = 0$. Rearranging this expression directly gives the value for the potential difference $V_A - V_B$ as +9 V.

The operating current of a device in a circuit is given by the formula $I = \frac{P}{V}$. Using the rated power of 500 W and voltage of 100 V, the current is calculated as 5 A. For the bulb to function at its rated power in a 230 V supply, the excess voltage of 130 V must drop across the series resistor. Applying Ohm's law, the required resistance is found by dividing this voltage drop by the 5 A current, resulting in 26 ohms.

The total heat generated in a resistor is found by integrating the rate of thermal energy production, which is governed by the relation $H = \int i^2 R \, dt$. The instantaneous current through the circuit is derived by taking the time derivative of the charge function, resulting in $i = \frac{dQ}{dt} = a - 2bt$. Current continues to flow until it drops to zero, determining the total duration of heat generation as $t = \frac{a}{2b}$. Substituting the current equation into the heat integral and evaluating it from time zero to this upper limit yields the accumulated thermal energy, which ultimately simplifies to $\frac{a^3R}{6b}$.

The internal resistance of an ammeter containing a galvanometer coil and a parallel shunt resistor is calculated using the parallel resistance formula $R_A = \frac{R_{coil} R_{shunt}}{R_{coil} + R_{shunt}}$, which evaluates to 19.2 ohms. Adding this effective ammeter resistance to the external series resistor gives a total circuit resistance of 60 ohms. Applying Ohm's law with the 30 V battery source, the total current flowing through the circuit and registered by the ammeter is calculated as $30 / 60 = 0.5$ amperes.

The electrical resistance of a uniform wire is determined by its dimensions and material conductivity through the formula $R = \frac{l}{\sigma A}$. When two wires of identical length and cross-sectional area are connected in series, their combined resistance is the sum of their individual resistances, written as $R_{eq} = \frac{l}{A} (\frac{1}{\sigma_1} + \frac{1}{\sigma_2})$. Equating this sum to the equivalent series resistance formula containing the total length $2l$ and equivalent conductivity $\sigma_{eq}$ allows the cross-sectional area and length terms to cancel out. Rearranging the remaining expression directly yields the harmonic-mean-like combination formula for conductivity, $\frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$.

The potential gradient along a potentiometer wire depends on the total current circulating through the circuit and the resistance distributed uniformly across its length. First, determine the main current using Ohm's law as $I = \frac{E_0}{r + r_1}$, which allows calculation of the total potential drop across the wire itself as $V = I r = \frac{E_0 r}{r + r_1}$. Dividing this potential drop by the total length $L$ yields the potential gradient $k = \frac{V}{L}$. Since the balancing length is $l$, multiplying this gradient by $l$ gives the unknown emf as $E = k l = \frac{E_0 r}{r + r_1} \cdot \frac{l}{L}$.

In any electrical circuit, voltage division in series components depends directly on their resistance according to Ohm's law, while parallel branches always share the exact same potential difference. To determine the voltmeter readings, first calculate the equivalent resistance of voltmeters B and C connected in parallel using the equation $R_{BC} = \frac{1.5R \times 3R}{1.5R + 3R} = R$. Since voltmeter A has a resistance of $R$, it matches this parallel equivalent resistance. In the series arrangement, the voltage splits equally between voltmeter A and the parallel pair because their resistances are equal, establishing that $V_A = V_{BC}$. Furthermore, since voltmeters B and C are connected \in parallel, the potential difference across each is identical, resulting \in $V_B = V_C$, and ultimately leading to $V_A = V_B = V_C$.

A metallic conductor of varying cross-sectional area can be visualized as a continuous series combination of microscopic elemental resistors of different dimensions. In any series circuit configuration, the rate of flow of charge must remain uniform across all sections to prevent charge accumulation at any point. Consequently, the electric current passing through every cross-section of the conductor is identical, whereas quantities like drift velocity, electric field, and current density vary inversely with the local cross-sectional area.

The potential gradient along a potentiometer wire is defined as the potential drop per unit length, expressed through the relation $k = \frac{V_{wire}}{L}$. Converting the given gradient of 1 mV per cm yields 0.1 V per meter, which for a 4-meter wire requires a total potential drop of 0.4 V. The total current in the primary circuit is governed by the expression $I = \frac{V}{R_{total}} = \frac{2}{8 + R}$, creating a potential drop across the wire equal to $\frac{16}{8 + R}$. Equating the resulting gradient to the required value allows us to solve for the series resistance $R$, yielding 32 ohms.

Power loss in transmission wires is calculated using the Joule heating formula $P = I^2 R$, which requires finding the total resistance and current of the entire line. The total resistance is obtained by multiplying the total distance by the resistance per unit length, giving $150 \times 0.5 = 75$ ohms. The current is determined by dividing the potential drop per unit length by the resistance per unit length, yielding $I = 8 / 0.5 = 16$ amperes. Squaring this current and multiplying it by the total resistance results in $P = (16)^2 \times 75 = 19200$ watts, which simplifies to $19.2$ kW.

The internal resistance of a cell is related to the balancing lengths obtained on a potentiometer wire with and without an external shunt resistance through the formula $r = R \left(\frac{l_1}{l_2} - 1\right)$. When the external resistance is infinite, the balancing length corresponds to the true emf of the cell, giving $l_1 = 3$ meters. When the resistance of $9.5$ ohms is connected, the terminal potential difference is balanced at $l_2 = 2.85$ meters. Substituting these values into the formula yields $r = 9.5 \times \left(\frac{3}{2.85} - 1\right)$, which simplifies to $9.5 \times \left(\frac{0.15}{2.85}\right)$, resulting in an internal resistance of $0.5$ ohms.