Current Electricity – NEET UG PhysicsPractice Questions & PYQs

A meter bridge functions based on the Wheatstone bridge principle, establishing that the ratio of two resistances equals the ratio of their corresponding balancing lengths on the meter wire, expressed as $\frac{P}{Q} = \frac{l}{100-l}$. For the initial setup with resistances of 5 ohms and $R$ ohms balancing at a length $l_1$, this relationship gives the equation $\frac{5}{l_1} = \frac{R}{100-l_1}$. When the resistance $R$ is shunted with an identical resistor, the combined resistance becomes half of its original value, shifting the new balance point to $1.6l_1$. Incorporating these updated values into the bridge formula yields the second condition as $\frac{5}{1.6l_1} = \frac{R}{2(100-1.6l_1)}$. Comparing these two relations allows us to solve for the initial balancing length, resulting in $l_1 = 25$, and substituting this value back into the first expression finally reveals the resistance $R$ to be 15 ohms.

The resistance of a uniform wire segment is directly proportional to its length, meaning the individual sides possess baseline resistances proportional to 3, 4, and 5. When a multimeter measures the resistance across any two vertices, that specific side acts as a single branch connected in parallel with the remaining two sides combined in series.

For the first pair of terminals, the side of length 3 sits in parallel with the series sum of lengths 4 and 5, producing an equivalent resistance calculated via $R_{AB} = \frac{R_1(R_2 + R_3)}{R_1 + R_2 + R_3}$ which evaluates to $\frac{27}{12}$ multiplied by the material and geometric constants. Applying this exact circuit configuration to the second and third pairs of terminals results in equivalent resistances of $\frac{32}{12}$ and $\frac{35}{12}$ respectively. Evaluating the ratio of these three measurements eliminates the common factors and yields 27:32:35.

The resistance of a conductor depends on its dimensions through the relation $R = \rho \frac{l}{A}$. Because the total volume of the wire stays constant when stretched, doubling the length ($l' = 2l$) forces the cross-sectional area to decrease by half ($A' = A/2$). Applying these changes to the resistance equation results in $R' = \rho \frac{2l}{A/2} = 4 \left(\rho \frac{l}{A}\right)$, which yields a final resistance of 16 ohms when multiplied by the initial 4 ohm value.

To find the required voltage, we apply Ohm's law using the equation $V = I R$, combining the individual resistances of the two rods connected in series. We first convert the given values into consistent SI units, where the cross-sectional area is 0.01 cm^2 or 0.01 x 10^-4 m^2, the length of each rod is 1 m, and the converted resistivities are 1.7 x 10^-8 ohm-meters for copper and 10^-7 ohm-meters for iron. Applying the resistance formula $R = \rho \frac{l}{A}$, we calculate the copper rod's resistance as 0.017 ohms and the iron rod's resistance as 0.1 ohms. Summing these series resistances gives a total resistance of 0.117 ohms, which, when multiplied by the current of 1 ampere, yields a voltage of 0.117 volts.

The relationship between a cell's electromotive force, current, external load, and internal resistance is expressed by the energy conservation equation $\varepsilon = I(R + r)$. Substituting the known values of emf as $2.1$ volts, current as $0.2$ amperes, and external resistance as $10$ ohms into this formula gives $2.1 = 0.2 \times (10 + r)$. Expanding the right side yields $2.1 = 2 + 0.2r$, which rearranges to $0.2r = 0.1$. Solving for the internal resistance results in $r = 0.5$ ohms.

A Wheatstone bridge achieves a balanced condition when the ratio of adjacent arm resistances satisfies $\frac{P}{Q} = \frac{R}{S}$. Checking the given values gives $\frac{10}{30} = \frac{30}{90}$, confirming that the bridge is balanced because both ratios simplify to $\frac{1}{3}$. Consequently, no current flows through the galvanometer, making the $50$ ohm galvanometer resistance completely inactive. The total equivalent resistance of the circuit simplifies to the internal resistance of the cell in series with the parallel-series combination of the bridge arms, calculated as $5 + \frac{(10 + 30)(30 + 90)}{(10 + 30) + (30 + 90)} = 5 + \frac{40 \times 120}{160} = 5 + 30 = 35$ ohms. Applying Ohm's law to the entire circuit, the current drawn from the cell is $I = \frac{7}{35} = 0.2$ amperes.

When a galvanometer shows no deflection, it indicates that zero current passes through that branch, meaning the potential difference across the parallel resistor balances the emf of the connected cell. The main loop current driven by cell A through resistor $R_1$ and resistor $R$ is calculated using the formula $I = \frac{V_A}{R_1 + R}$. Substituting the given values into this relation yields $I = \frac{12}{500 + 100}$, which simplifies to 12/600 amperes. Since the galvanometer branch carries no current, the voltage of cell B must equal the potential drop across resistor $R$. Using the relation $V_B = I \times R$, multiplying the current by 100 ohms gives a potential of 2 volts.

When a current enters a ring split into two parts, the total resistance of 12 ohms is divided into two parallel branches whose resistances depend directly on their respective lengths, $l_1$ and $l_2$. The equivalent resistance between connection points A and B is determined by the parallel resistance formula $R = \frac{R_1 R_2}{R_1 + R_2}$, which is given as 8/3 ohms. Since the sum of the individual branch resistances equals the total resistance of 12 ohms, each branch resistance can be scaled by the length ratio of the segments. Substituting these proportional relationships into the parallel resistance expression sets up a quadratic equation in terms of the length ratio. Solving this quadratic relation yields possible values of 2 and 1/2, establishing that the required ratio of the lengths is $\frac{l_1}{l_2} = \frac{1}{2}$.

The electrical power of a device operating at a constant resistance is related to the applied voltage by the formula $P = \frac{V^2}{R}$. Taking the differential form of this relation shows that the relative percentage change in power is twice the relative percentage change in voltage. Substituting the given 2.5% voltage drop into this proportional relationship gives a total power decrease of 5%.

The relationship between electrical power, voltage, and equivalent resistance is expressed by the formula $P = \frac{V^2}{R_{\text{eq}}}$ where power is equal to the square of the applied voltage divided by the total resistance of the circuit.

Using the provided power of 30 watts and a source voltage of 10 volts, along with the network equivalent resistance of $\frac{5R}{5+R}$, the equation becomes $30 = \frac{10^2}{\frac{5R}{5+R}}$ Simplifying the numerator gives $100$, and transferring the complex denominator yields $30 × \frac{5R}{5+R} = 100$. Multiplying both sides by $(5+R)$ transforms the expression into $150R = 100(5+R)$, which expands to $150R = 500 + 100R$. Collecting the variable terms on one side results \in $50R = 500$ and dividing by 50 successfully isolates the resistance value to 10 ohms.

The potential difference across a variable external resistance connected to a source is governed by the equation $V = \frac{\varepsilon R}{R+r}$, which describes how the total voltage divides across the circuit components. When the external resistance is reduced to zero, the potential difference across it is likewise zero. As the external resistance is continuously increased toward infinity, the voltage rises in a gradual curve and asymptotically approaches the maximum emf value of the cell.

Finding the potential at a specific point in a circuit requires applying Kirchhoff's Voltage Law, which tracks the cumulative changes in electric potential along a chosen path. Moving from point A to point B along the circuit branch, the sum of the initial potential and all potential variations is expressed by the equation $V_A + 1 + 2(1) - 2 = V_B$. Given that the potential at point A is zero, substituting this value yields $0 + 1 = V_B$, which calculates out to a final potential of +1 V at point B.

The relationship between a battery's electromotive force ($\varepsilon$), internal resistance ($r$), external load ($R$), and current ($I$) is expressed as $I = \frac{\varepsilon}{R + r}$. Setting up this equation for the two given scenarios yields $2 = \frac{\varepsilon}{2 + r}$ and $0.5 = \frac{\varepsilon}{9 + r}$. Dividing the first equation by the second eliminates the emf and allows for a direct algebraic solution for $r$, resulting in an internal resistance of one-third of an ohm.

The power dissipated in a resistor is expressed by the relation $P = I^2R$. Using this relationship, the current flowing through the 9 ohm resistor is found by dividing the given power by the resistance and taking the square root, which results in 2 amperes. Since the 9 ohm and 6 ohm resistors are connected in parallel, the potential difference across them is identical, allowing us to determine that the current through the 6 ohm resistor is 3 amperes. Adding these two branch currents together gives a total current of 5 amperes drawn from the source. Multiplying this total current by the 2 ohm resistance yields a potential difference of 10 volts.

A potentiometer determines unknown potential differences by equating them to the potential drop across a calibrated wire segment, defined as the product of the potential gradient and the balancing length. When the key is connected between terminals 1 and 2, the balancing length $l_1$ measures the potential drop solely across resistor $R$. Given that the circuit current is 1.0 A and the potential gradient is $k$, the equation $IR = kl_1$ directly yields $R = kl_1$.

When the key shifts to connect terminals 1 and 3, the balancing length $l_2$ accounts for the total potential drop across both resistors in series, represented by $I(R + X) = kl_2$. Substituting the current of 1.0 A into this expression gives $R + X = kl_2$. Subtracting the value of $R$ from this total resistance reveals that $X$ equals $k(l_2 - l_1)$, completing the derivation for both resistor values.

Kirchhoff's first law relies on the fact that electric charge cannot accumulate at a node, thereby reflecting the fundamental principle of conservation of charge. Similarly, Kirchhoff's second law dictates that the total potential drop around a closed circuit loop equals the supplied source potential, which embodies the law of conservation of energy. Consequently, both statements accurately describe the physical laws underlying electrical circuit analysis.

Kirchhoff's loop rule states that the algebraic sum of potential changes around any closed electrical circuit path is always zero. Traversing the designated loop in the direction of the currents causes a potential drop of $-({i_1}+{i_2}){R}$ across the external load and another drop of $-{i_1}{r_1}$ across the internal resistance. Moving across the battery from its negative to its positive terminal provides a potential rise of $+{\varepsilon_1}$. Combining these potential changes into a balanced loop equation produces ${\varepsilon_1}-({i_1}+{i_2}){R}-{i_1}{r_1}=0$.

The energy gained by an electron moving through an electric field is equal to the product of its charge, the field strength, and the distance traveled, expressed by the relation $W = e E \lambda$. Equating the given energy of 2 eV to $e E \lambda$ cancels the elementary charge, simplifying the expression to $2 = E \lambda$. Substituting the mean free path value of $4 \times 10^{-8} \ \mathrm{m}$ into this rearranged equation yields $E = \frac{2}{4 \times 10^{-8}}$, which calculates directly to $5 \times 10^{7} \ \mathrm{V/m}$.

The terminal potential difference of a real cell varies with current according to the equation $V = \varepsilon - Ir$. Rearranging this into standard linear form yields $V = -rI + \varepsilon$, where terminal voltage acts as the dependent variable and current as the independent variable. In this graphical representation, the constant term serves as the maximum potential intercept at zero current, while the coefficient of current represents the negative slope equal to $-r$.

The equivalent resistance of a divided circular wire is determined by treating its semicircular halves as resistors connected in parallel. The total length of the wire forming the circular loop is equal to its circumference, calculated as two times pi times the radius of 10 cm to give 0.2 pi meters. Multiplying this length by the resistance rate of 12 ohms per meter yields a total wire resistance of 2.4 pi ohms. Because the connection points are diametrically opposite, they divide the loop into two identical semicircular branches, each possessing a resistance of 1.2 pi ohms. Substituting these values into the parallel resistance formula, $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$, establishes that the combined resistance between the points is 0.6 pi ohms.