NEET Current Electricity Previous Year Questions with Solutions

In series, the total equivalent resistance is the algebraic sum of the individual resistances:

• • The linear temperature-dependent terms (-alpha*Delta T and +alpha*Delta T) cancel out completely.
• • R_{eq,A} = 2 R_0 is completely independent of temperature.
• • Therefore, current I_A = V / (2 R_0) remains strictly constant as temperature increases.

In parallel across constant voltage V, total current is the sum of branch currents:

• • Find the common denominator: (1 - alpha*Delta T)(1 + alpha*Delta T) = 1 - alpha^2 (Delta T)^2.
• • I_B = \frac{V}{R_0} \left[ \frac{(1 + \alpha \Delta T) + (1 - \alpha \Delta T)}{1 - \alpha^2 (\Delta T)^2} \right] = \frac{2V}{R_0 \left[1 - \alpha^2 (\Delta T)^2\right]}.

Examine the behavior of the denominator as Delta T increases:

• • Because the denominator decreases below 1, total parallel current I_B strictly INCREASES.
• • Conclusion: I_A remains constant while I_B increases.
• • Matches Option (1) exactly.

In the circuit diagram, line AF is a plain conducting wire, which implies V_A = V_F. Similarly, line CD is a plain conducting wire, meaning V_C = V_D. Redrawing the internal mesh between nodes B and E:

• • Resistor R_FE (3 Ω) is effectively connected between node A and node E.
• • Resistor R_BC (2.5 Ω) is effectively connected between node B and node D.
• • Resistor of 6 Ω is connected diagonally across node A and node D.

Examine the ratios of the arms connected between terminals B and E:

• • Because the ratios are equal, potential at node A equals potential at node D (V_A = V_D).
• • No current flows through the 6 Ω bridge resistor; it can be removed from the circuit.

With the 6 Ω resistor removed, the upper series arm (5 + 3 = 8 Ω) is in parallel with the lower series arm (2.5 + 1.5 = 4 Ω):

• • The external circuit contains series resistors: (1.5 + 5.5) = 7 Ω and (1/3) Ω.
• • Total resistance: R_total = R_BE + 7 + 1/3 = 8/3 + 1/3 + 7 = 3 + 7 = 10 Ω.
• • Total current through battery: I = V / R_total = 5 V / 10 Ω = 0.5 A.

When a uniform wire of total resistance R is cut into 8 equal lengths, the cross-sectional area and resistivity remain unchanged. Therefore, each piece has resistance:

• • Linear dependence: R ∝ L.

Four pieces of resistance r are connected in parallel. For n identical resistors in parallel, the equivalent resistance is r / n:

• • Both sets are identical, so each set has resistance R / 32.

Connecting the two parallel sets in series gives a net resistance equal to the sum of the two resistances:

• • Matches Option (3) exactly.

Branch CD is an ideal wire with zero resistance connecting junction C and junction D. Therefore, junctions C and D form a single equipotential node:

• • The 1 Ω and 3 Ω resistors are connected in parallel between node A and node CD.
• • The 2 Ω and 4 Ω resistors are connected in parallel between node CD and node B.

Calculate the equivalent resistance of each section and the whole network:

• • Total current delivered by the 50 V battery:
• • I_{total} = \frac{50}{25/12} = 50 \times \frac{12}{25} = 24\text{ A}.

Determine the potential at the central node and find currents entering and leaving node C:

• • V_C - V_B = 50 - 18 = 32 V ⟹ I_{CB} = 32 V / 2 Ω = 16 A.
• • KCL at node C: I_{AC} = I_{CB} + I_{CD} ⟹ I_{CD} = 18 A - 16 A = 2.0 A (flowing downward from C to D).
• • Verification at node D: I_{AD} = 18 V / 3 Ω = 6 A; I_{DB} = 32 V / 4 Ω = 8 A; 6 A + 2 A = 8 A (verified!).

Since resistance of a uniform wire is directly proportional to its length (R = ρl/A), dividing the wire into 10 equal parts divides its total resistance equally among the 10 segments:

The first 5 parts are connected end-to-end in series. For series combination, individual resistances simply add up:

The next 5 parts are connected in parallel across common terminals. For 5 identical resistors in parallel:

The problem states that these two combinations are now connected in series with each other. Therefore, the total equivalent resistance is:

According to Ohm’s law applied to the complete closed circuit consisting of the real battery and the external resistor, the current is:

During discharge through an external load, the terminal voltage V across the battery terminals is reduced from its open-circuit EMF by the internal voltage drop Ir:

The terminal voltage must also identically equal the potential difference across the external 4 Ω resistor:

For a heater designed for rated voltage V, power is P = V² / R, which gives resistance R = V² / P:

When connected in series across the same fixed voltage source V, the total equivalent resistance is:

The power delivered in the series configuration is:

When connected in parallel across the same fixed voltage V, each heater has the full source voltage across it. Power outputs add directly:

Taking the ratio of the series power to parallel power:

When the galvanometer G shows zero deflection, the current flowing through its branch is zero (i_g = 0). By Kirchhoff's Current Law at the junction, the secondary branch is completely isolated in terms of current flow:

All current from the 10 V source circulates exclusively through the series combination of 400 Ω and R:

The potential drop across resistor R is given by Ohm's law and must balance the 2 V battery:

Solve the linear equation for resistance R:

Standard 4-band carbon resistors encode resistance as R = (AB × 10^C ± D%) Ω, where A and B are the first two significant figures, C is the decimal multiplier, and D is the tolerance percentage:

Comparing with the standard colour-code format:

Using the canonical mnemonic "B B R O Y of Great Britain has Very Good Wife":

The given circuit is a single closed loop ABCDA. All components (the 2 Ω resistor, the 10 V cell, junction E, the 5 V cell, the 1 Ω resistor, and the 7 Ω resistor) are connected in series.

The positive terminal of the 10 V source faces towards E (pushing current from left to right, i.e., from A towards B), while the 5 V source faces into E with its positive terminal pushing opposite. Because they are in opposition, the net driving EMF is the difference:

By Ohm’s Law / Kirchhoff’s Loop Rule applied to the single closed path:

Since E₁ (10 V) is greater than E₂ (5 V), the net current flows in the direction driven by the 10 V battery: flowing clockwise through the loop, which corresponds to travelling from A to B through point E.

When 10 identical resistors of resistance R are connected in series, the equivalent resistance R_s is the direct sum of all individual resistances:

Connected across an ideal battery of EMF E with negligible internal resistance, the circuit current I_s is given by Ohm’s Law:

When the same 10 resistors are connected in parallel, the reciprocal sum yields equivalent resistance R_p:

Connected across the same battery of EMF E, the circuit current I_p is:

The problem states that I_p = n I_s. Taking the ratio:

Over moderate temperature intervals, the electrical resistance of a metallic conductor varies linearly with temperature according to the relation:

From the problem statement:

Expanding and isolating the temperature coefficient α:

Substitute the given values into the isolated equation:

Convert 0.03 to scientific notation:

Under an applied electric field $\vec{E}$, the electric force on an electron is $-e\vec{E}$, imparting acceleration $\vec{a} = -e\vec{E}/m$. The average velocity acquired over relaxation time $\tau$ is:

In vector microscopic notation, Ohm's law states $\vec{J} = \sigma \vec{E} = \frac{1}{\rho}\vec{E}$. Rearranging gives electrical resistivity:

Electric current is $I = n e A v_d$. Dividing by cross-sectional area $A$ gives the current density:

Microscopic resistivity in terms of electron mass $m$, charge $e$, number density $n$, and relaxation time $\tau$ is $\rho = \frac{m}{n e^2 \tau}$. Solving for $\tau$ gives:

Since the four wires have identical length $l$, area $A$, and resistivity $\rho$, each wire has the same resistance $R = \rho l / A$. When connected in parallel, their equivalent resistance $R_p$ is:

Given $R_p = 0.25\ \Omega$:

When the four identical wires of resistance $R = 1.0\ \Omega$ are connected in series:

For a uniform potentiometer wire carrying a steady current, the potential drop per unit length (potential gradient $k$) is constant. At null deflection, the EMF of the test cell is balanced by the potential drop across balancing length $l$:

Comparing the two test cells balanced on the same potentiometer wire:

Substitute $l_1 = 36\text{ cm}$, $E_1 = 1.5\text{ V}$, and $E_2 = 2.5\text{ V}$:

Current $i_1$ enters from terminal A and passes entirely through resistor $r_1$. At the junction ahead of $r_1$, the current divides into two parallel branches containing $r_2$ and $r_3$:

Because resistors $r_2$ and $r_3$ are connected across the exact same pair of nodes, their potential drops are equal:

Substitute $i_2 = i_1 - i_3$ into the voltage equality equation:

According to the Drude-Lorentz free electron model, the electrical resistivity of a solid is given by: $$\rho = \frac{m}{n e^2 \tau}$$ where $$m$$ is electron mass, $$e$$ is electron charge, $$n$$ is charge carrier number density, and $$\tau$$ is the average relaxation time between collisions.

In metals, the carrier concentration $$n$$ is independent of temperature ($$\sim 10^{28}\text{ m}^{-3}$$). As temperature increases, lattice thermal vibrations amplify, reducing relaxation time $$\tau$$. Consequently, $$\rho$$ increases with temperature, yielding a POSITIVE temperature coefficient of resistance ($$\alpha > 0$$).

In semiconductors and insulators, valence electrons are separated from the conduction band by an energy band gap ($$E_g$$). As temperature rises, thermal energy excites electrons across the band gap into the conduction band, causing the carrier density $$n$$ to increase exponentially according to Boltzmann distribution: $$n(T) \propto e^{-E_g / (2 k_B T)}$$ This immense increase in $$n$$ overwhelmingly compensates for any modest decrease in relaxation time $$\tau$$. Consequently, resistivity and resistance decrease exponentially as temperature increases.

Because $$\frac{dR}{dT} < 0$$, both insulators and semiconductors possess a NEGATIVE temperature coefficient of resistance ($$\alpha < 0$$).