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

Q jee_main_2026_21_jan_morning Heating Effect of Current

The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9V with internal resistance of

1 Ω

is connected across these points is ____ J.

Heating Effect of Current diagram for Q47 - JEE Main 2026 Morning — A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.

Numerical Answer. Answer: 1080 to 1080

Solution

Related Formula

Req = (R₁ R₂)/(R₁ + R₂) (parallel)

$H = I² R t$ $H = I^2 R t$

I = VReq + r

Core Logic

The circuit is a balanced Wheatstone bridge between A and B, because the ratio of adjacent arms is $(1)/(2) = (2)/(4)$ $\frac{1}{2} = \frac{2}{4}$. Thus, the middle $1 Ω$ $1\,\Omega$ resistor (if there is one connecting the middle nodes) is ineffective.

Heating Effect of Current solution diagram for Q47 - JEE Main 2026 Morning — A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.

The equivalent resistance across AB ( $RAB$ $R_{AB}$) involves the upper branch ( $1+2 = 3 Ω$ $1+2 = 3\,\Omega$) and the lower branch ( $2+4 = 6 Ω$ $2+4 = 6\,\Omega$) in parallel.

RAB = (3 \times 6)/(3 + 6) = (18)/(9) = 2 Ω

Step 1: Calculate Current and Heat

Total resistance in the circuit = $RAB + rinternal = 2 + 1 = 3 Ω$ . Total current from the battery $i = 9V3 Ω = 3A$ .

The heat generated specifically between points A and B in 1 minute (60 seconds) is:

HAB = i² RAB t = (3)² \times 2 \times 60

HAB = 9 \times 120 = 1080 J

Pattern Recognition

Check for balanced Wheatstone bridge first. Then use $H = i² R t$ $H = i^2 R t$ strictly with the equivalent resistance of just the section $AB$ $AB$ to find heat specifically generated across $AB$ $AB$.

Chapter Mix

Class 12 Physics: Current Electricity

Q28 jee_main_2026_21_jan_evening Potentiometer

The total length of potentiometer wire AB is

50 cm

in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is ________ cm.

Potentiometer diagram for Q28 - JEE Main 2026 Evening — Potentiometer circuit with a 50 cm wire AB, a galvanometer, and standard bridge resistors of 6 ohms and 4 ohms.

A. $15$ $15$
B. $30$ $30$
C. $25$ $25$
D. $20$ $20$

Solution

Related Formula

For a balanced Wheatstone bridge configuration:

(R₁)/(R₂) = (l₁)/(l₂)

Core Logic

When the galvanometer shows zero reading, it implies that the bridge is balanced. The resistance of the wire segments is proportional to their lengths.

Let the length of segment AP be $lAP$ $l_{AP}$ and PB be $lPB$ $l_{PB}$.

6 ΩRAP = 4 ΩRPB

Since $R \propto l$ $R \propto l$:

RAPRPB = lAPlPB = (6)/(4) = (3)/(2)

Step 1: Applying Total Length Constraint

We are given the total length:

lAP + lPB = 50 cm

Using the ratio $lAP = (3)/(2) lPB$ $l_{AP} = \frac{3}{2} l_{PB}$:

(3)/(2)lPB + lPB = 50

(5)/(2)lPB = 50 lPB = 20 cm

Step 2: Final Conclusion

Solving for $lAP$ $l_{AP}$:

lAP = 50 - 20 = 30 cm

Alternatively, $lAP = (3)/(3+2) \times 50 = (3)/(5) \times 50 = 30 cm$ .

Pattern Recognition

Null deflection over a uniform wire directly translates resistance ratios into length ratios. Standard Wheatstone bridge application wrapped in potentiometer geometry.

Chapter Mix

Class 12 Physics: Current Electricity

Q35 jee_main_2026_21_jan_evening Maximum Power Transfer Theorem

A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when :

A. $R=2r$ $R=2r$
B. $R=(r)/(2)$ $R=\frac{r}{2}$
C. $R=\sqrt(2)r$ $R=\sqrt{2}r$
D. $R=r$ $R=r$

Solution

Related Formula

P = I² R = ((E)/(R+r))² R

Core Logic

To maximize power across the load resistance $R$ $R$, we apply the Maximum Power Transfer Theorem. The power transferred to the load is maximum when the load resistance equals the internal resistance of the battery.

Setting $(dP)/(dR) = 0$ $\frac{dP}{dR} = 0$ yields: $R = r$ $R = r$

Step 1: Final Conclusion

Power is maximum when the external resistance is strictly equal to the internal resistance ( $R=r$ $R=r$).

Pattern Recognition

Direct theoretical application of the Maximum Power Transfer Theorem.

Chapter Mix

Class 12 Physics: Current Electricity

Q40 jee_main_2026_21_jan_evening Combination of Resistors

Two known resistance of

R Ω

and

2R Ω

and one unknown resistance

X Ω

are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is

X Ω

, then the value of X is ________

Ω

Circuit diagram of resistors for Q40 - JEE Main 2026 Evening — Bridge-like resistor circuit containing unknown resistance X and known resistances R and 2R.

A. $(\sqrt(3)-1)R$ $(\sqrt{3}-1)R$
B. $R$ $R$
C. $2(\sqrt(3)-1)R$ $2(\sqrt{3}-1)R$
D. $(\sqrt(3)+1)R$ $(\sqrt{3}+1)R$

Solution

Related Formula

For parallel combination: $Req = (R₁ R₂)/(R₁ + R₂)$ For series combination: $Req = R₁ + R₂$ $R_{\text{eq}} = R_1 + R_2$

Core Logic

Based on the circuit topology, the $2R$ $2R$ and $X$ $X$ resistors are in series. This combination is in parallel with the $R$ $R$ resistor. The total equivalent resistance across A and B is given as $X$ $X$.

Req = ((2R + X) \cdot (R))/((2R + X) + R)

Step 1: Setting up the Quadratic

Set the equivalent resistance equal to $X$ $X$:

X = ((2R + X)R)/(3R + X)

Cross-multiplying yields:

X(3R + X) = 2R² + XR

3RX + X² = 2R² + XR

X² + 2RX - 2R² = 0

Step 2: Solving for X

Using the quadratic formula to solve for $X$ $X$:

X = -2R \pm \sqrt((2R)² - 4(1)(-2R²))2

X = -2R \pm \sqrt(4R² + 8R²)2 = -2R \pm \sqrt(12R²)2

X = -2R \pm 2R\sqrt(3)2 = -R \pm R\sqrt(3)

Since resistance must be positive:

X = (\sqrt(3) - 1)R

Pattern Recognition

Self-referential equivalent resistance problems always collapse into a quadratic equation $x² + ax + b = 0$ $x^2 + ax + b = 0$. Solve strictly for the positive root.

Chapter Mix

Class 12 Physics: Current Electricity

Q31 jee_main_2026_22_january_morning Meter Bridge

A meter bridge with two resistances

R₁

and

R₂

as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when

16~Ω

resistance is connected in parallel to

R₂

. The values of resistances

R₁

and

R₂

are \_\_\_\_.

Current Electricity diagram for Q31 - JEE Main 2026 January Morning — Meter bridge circuit with resistances R1 and R2 and balancing lengths.

A. $R₂ = 16Ω, R₁ = (16)/(3)Ω$
B. $R₂ = 4Ω, R₁ = (4)/(3)Ω$
C. $R₂ = 8Ω, R₁ = (16)/(3)Ω$
D. $R₂ = 12Ω, R₁ = (12)/(3)Ω$

Solution

Related Formula

(R₁)/(R₂) = (l₁)/(100 - l₁)

Core Logic

Solution diagram for Q31 - JEE Main 2026 Morning — Meter bridge circuit with resistances R1 and R2 and balancing lengths.

Initially:

$(R₁)/(R₂) = (40)/(60) = (2)/(3) R₁ = (2)/(3)R₂$ ---- (1)

After connecting $16~Ω$ $16~\Omega$ in parallel with $R₂$ $R_2$:

$(R₁)/(((R₂ \times 16)/(R₂ + 16))) = (50)/(50) R₁ = (16 R₂)/(16 + R₂)$ ---- (2)

Solving equations (1) and (2) yields:

R₂ = 8~Ω, R₁ = (16)/(3)~Ω

Pattern Recognition

Sees: Meter bridge balancing with parallel shunt resistance. Shortcut: Set up length ratios for initial and modified states and solve linear simultaneous equations. Check: Matches option (3). ✓

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

Solution

Related Formula

Core Logic

Step 1: Calculate Current and Heat

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Total Length Constraint

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the Quadratic

Step 2: Solving for X

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

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.