A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The…

Year	2026	2025	2024	Total
Questions	18	13	19	50

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

For a wire of total resistance $R$ $R$ divided into $N$ $N$ equal segments, the resistance of each segment $r$ $r$ is:

r = (R)/(N)

For a balanced Wheatstone bridge with resistors of resistance $r$ $r$, the central arm can be neglected because no current flows through it.

Core Logic

The triangular pyramid has $6$ $6$ segments of equal length. Since the total resistance of the wire is $R$ $R$:

r = (R)/(6)

Let the four vertices of the pyramid be $A$ $A$, $B$ $B$, $C$ $C$, and $D$ $D$. Terminals are at $A$ $A$ and $B$ $B$.

The segments are:

$AB$ $AB$ (direct path between terminals, resistance $r$ $r$)
$AC$ $AC$, $BC$ $BC$, $AD$ $AD$, $BD$ $BD$ (forming a closed quadrilateral network between $A$ $A$ and $B$ $B$ with bridge arm $CD$ $CD$)
$CD$ $CD$ (bridge arm connecting the midpoints, resistance $r$ $r$)

Step 1: Simplify the Network

By symmetry, the potentials at $C$ $C$ and $D$ $D$ are equal when a voltage is applied across $A$ $A$ and $B$ $B$. Thus, the bridge is balanced, and no current flows through the segment $CD$ $CD$.

We can remove segment $CD$ $CD$ from the calculations:

The path $A \to C \to B$ $A \to C \to B$ consists of two resistors in series: $r + r = 2r$ $r + r = 2r$.
The path $A \to D \to B$ $A \to D \to B$ also consists of two resistors in series: $r + r = 2r$ $r + r = 2r$.
The direct path $A \to B$ $A \to B$ has a single resistor $r$ $r$.

These three parallel branches are connected between $A$ $A$ and $B$ $B$.

Step 2: Calculate Equivalent Resistance

The equivalent resistance $RAB$ $R_{AB}$ is:

1RAB = (1)/(2r) + (1)/(2r) + (1)/(r) = (1)/(r) + (1)/(r) = (2)/(r)

RAB = (r)/(2)

Substitute $r = (R)/(6)$ $r = \frac{R}{6}$:

RAB = (R/6)/(2) = (R)/(12)

Comparing with $RAB = (R)/(n)$ $R_{AB} = \frac{R}{n}$, we find $n = 12$ $n = 12$.

Pattern Recognition

Sees: Resistor network formed by a 3D pyramid (6 identical edges, 4 nodes). Shortcut: A 6-resistor regular tetrahedron has an equivalent resistance of $r/2$ $r/2$ across any two vertices. Since the total wire resistance is $R$ $R$ and it's cut into 6 pieces, $r = R/6 Req = R/12$ $r = R/6 \implies R_{\text{eq}} = R/12$.

Chapter Mix

Class 12 Physics: Current Electricity

Simplified bridge schematic for equivalent resistance of pyramid — A triangular pyramid resistor network with terminals A and B marked, showing symmetry in the layout.

More Current Electricity Previous-Year Questions — Page 5

Q1 jee_main_2025_29_jan_evening Seebeck Effect and Thermoelectricity

The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have:

A. $low thermal conductivity and low electrical conductivity$
B. $high thermal conductivity and high electrical conductivity$
C. $low thermal conductivity and high electrical conductivity$
D. $high thermal conductivity and low electrical conductivity$

Solution

Related Formula

V = S \cdot Δ T

where, $V$ $V$ = Thermoelectric voltage (Seebeck voltage) $S$ $S$ = Seebeck coefficient $Δ T$ $\Delta T$ = Temperature difference

Core Logic

To maximize the efficiency of a thermoelectric device harvesting heat energy, two conditions must be fulfilled:

Low thermal conductivity: This ensures that the temperature gradient ( $Δ T$ $\Delta T$) across the material is maintained and heat does not rapidly flow from the hot side to the cold side.
High electrical conductivity: This minimizes internal Joule heating losses ( $I²R$ $I^2R$) when electrical current is drawn from the material.

Therefore, the material should possess low thermal conductivity and high electrical conductivity.

Pattern Recognition

thermoelectric figure of merit is given by $Z = (S² σ)/(κ)$ $Z = \frac{S^2 \sigma}{\kappa}$, where $σ$ $\sigma$ is electrical conductivity and $κ$ $\kappa$ is thermal conductivity. To maximize $Z$ $Z$, we inherently need high $σ$ $\sigma$ and low $κ$ $\kappa$.

Chapter Mix

Class 12 Physics: Current Electricity Class 11 Physics: Thermal Properties of Matter

Q14 jee_main_2025_28_jan_morning Combination of Resistors

Find the equivalent resistance between two ends of the following circuit.

Combination of Resistors diagram for Q14 - JEE Main 2025 Morning — A complex looking resistive mesh whose node potentials reveal a simple symmetrical pattern.

A. $r$ $\mathrm{r}$
B. $r6$ $\frac{\mathrm{r}}{6}$
C. $r9$ $\frac{\mathrm{r}}{9}$
D. $r3$ $\frac{\mathrm{r}}{3}$

Solution

Core Logic

Label the circuit nodes carefully to see how the elements are connected across terminal zones:

Node potential labelling steps for Q14 — A complex looking resistive mesh whose node potentials reveal a simple symmetrical pattern.

Tracing parallel loops reveals all inner subsets share identical path boundaries.

Req = r/33 = r9

Step 1: Final Expression

The overall simplified system network value equals $r9$ $\frac{\mathrm{r}}{9}$, matching option (3).

Pattern Recognition

Short-circuit tracking lines allow collapsing complex meshes into basic parallel branches. If all nodes connect symmetrically, apply simple division: $R/N$ $\mathrm{R}/\mathrm{N}$.

Chapter Mix

Class 12 Physics: Current Electricity

Q15 jee_main_2025_28_jan_morning Combination of Resistors

A wire of resistance

R

is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

A. $9 / 8$ $9 / 8$
B. $8 / 9$ $8 / 9$
C. 27/32
D. 32/27

Solution

Related Formula

R = ρ A R \propto

Core Logic

For the equilateral triangle, the total resistance $R$ $\mathrm{R}$ is divided into three equal line parts of value $R3$ $\frac{\mathrm{R}}{3}$ each:

Visual representation of triangle and square edge divisions for Q15

An edge configuration places one part in parallel with the other two series steps:

(Req)₁ = ( 2R3) \cdot ( R3) 2R3 + R3 = (2)/(9)R

For the square, the wire resistance $R$ $\mathrm{R}$ divides into four parts of $R4$ $\frac{\mathrm{R}}{4}$ each:

An edge calculation pairs one section in parallel with the remaining three loops in series:

(Req)₂ = ( 3R4) \cdot ( R4) 3R4 + R4 = (3)/(16)R

Taking the ratio between these geometric layouts:

Step 1: Final Ratio Calculation

(Req)₁(Req)₂ = (2)/(9)R(3)/(16)R = (32)/(27)

This selects option (4).

Pattern Recognition

For a regular polygon with $N$ $\mathrm{N}$ identical sides formed from a wire of resistance $R$ $\mathrm{R}$, the resistance across one edge is always given by $(N-1)N²R$ .

Chapter Mix

Class 12 Physics: Current Electricity

Q5 jee_main_2025_03_april_morning Resistance of a Circular Wire

A wire of length

25~m

and cross-sectional area

5~mm²

having resistivity of

2\times 10⁻⁶Ω~m

is bent into a complete circle. The resistance between diametrically opposite points will be:

A. $12.5Ω$ $12.5\Omega$
B. $50Ω$ $50\Omega$
C. $100Ω$ $100\Omega$
D. $25Ω$ $25\Omega$

Solution

Related Formula

Resistance of a uniform wire:

R = (ρ L)/(A)

Equivalent resistance of two identical resistors in parallel:

Req = (R₁ R₂)/(R₁ + R₂) = Rhalf2

Core Logic

First, find the total resistance $Rtotal$ $R_{\text{total}}$ of the straight wire:

Length of wire, $L = 25~m$ $L = 25\mathrm{~m}$
Area of cross-section, $A = 5~mm² = 5 \times 10⁻⁶~m²$
Resistivity, $ρ = 2 \times 10⁻⁶Ω~m$

Rtotal = (ρ L)/(A) = 2 \times 10⁻⁶ \times 255 \times 10⁻⁶ = 10Ω

Step 1: Splitting into Semicircles

When the wire is bent into a complete circle, measuring the resistance between two diametrically opposite points splits the wire into two parallel halves of equal length.

Each semicircle has a resistance of:

Rsemi = Rtotal2 = (10)/(2) = 5Ω

These two halves are connected in parallel between the diametric terminals:

Req = Rsemi2 = (5)/(2) = 2.5Ω

Circular loop splitting resistance across diameter for Q5

Step 2: Conclusion & Discrepancy

The actual mathematically rigorous answer is $2.5Ω$ $2.5\Omega$. Since $2.5Ω$ $2.5\Omega$ is not present in the given options, the question is marked as a Bonus question by standard key evaluation guidelines. If forced to choose a theoretical option due to printing mistakes, some keys may relate it to $10Ω / 4 = 2.5Ω$ $10\Omega / 4 = 2.5\Omega$, but scientifically it stands as a bonus.

Pattern Recognition

Shortcut: A wire of total resistance $R$ $R$ bent into a circle has an effective resistance across its diameter equal to $Req = R/4$ $R_{\text{eq}} = R/4$. Memorize this ratio! Here $R = 10Ω$ $R = 10\Omega$, so $Req = 10/4 = 2.5Ω$ $R_{\text{eq}} = 10/4 = 2.5\Omega$.

Chapter Mix

Class 12 Physics: Current Electricity

A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points A and B is R/n. The value of n is : A triangular pyramid resistor network with terminals A and B marked, showing symmetry in the layout.

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify the Network

Step 2: Calculate Equivalent Resistance

Pattern Recognition

Chapter Mix

Reference Study Guides

More Current Electricity Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Ratio Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Splitting into Semicircles

Step 2: Conclusion & Discrepancy

Pattern Recognition

Chapter Mix

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A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points A and B is R/n. The value of n is :
A triangular pyramid resistor network with terminals A and B marked, showing symmetry in the layout.