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

For the equilateral triangle, the total resistance R is divided into three equal line parts of value R3 each: An edge configuration places one part in parallel with the other two series steps: For the square, the wire resistance R divides into four parts of R4 each: An edge calculation pairs one section in parallel with the remaining three loops in series: Taking the ratio between these geometric layouts: This selects option (4). For a regular polygon with N identical sides formed from a wire of resistance R, the resistance across one edge is always given by (N-1)N2R. Class 12 Physics: Current Electricity

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 sq

For the equilateral triangle, the total resistance R is divided into three equal line parts of value R3 each: An edge configuration places one part in parallel with the other two series steps: For the square, the wire resistance R divides into four parts of R4 each: An edge calculation pairs one section in parallel with the remaining three loops in series: Taking the ratio between these geometric layouts: This selects option (4). For a regular polygon with N identical sides formed from a wire of

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

For the equilateral triangle, the total resistance R is divided into three equal line parts of value R3 each: An edge configuration places one part in parallel with the other two series steps: For the square, the wire resistance R divides into four parts of R4 each: An edge calculation pairs one section in parallel with the remaining three loops in series: Taking the ratio between these geometric layouts: This selects option (4). For a regular polygon with N identical sides formed from a wire of resistance R, the resistance across one edge is always given by (N-1)N2R. Class 12 Physics: Current Electricity

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 sq

For the equilateral triangle, the total resistance R is divided into three equal line parts of value R3 each: An edge configuration places one part in parallel with the other two series steps: For the square, the wire resistance R divides into four parts of R4 each: An edge calculation pairs one section in parallel with the remaining three loops in series: Taking the ratio between these geometric layouts: This selects option (4). For a regular polygon with N identical sides formed from a wire of

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of…

Year	2026	2025	2024	Total
Questions	18	13	19	50

Q37 jee_main_2026_24_january_morning Meter Bridge

Two resistors

2Ω

and

3Ω

are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with

3Ω

resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor is ____

Ω

.

Meter bridge circuit diagram — Meter bridge with 2 ohm and 3 ohm resistors connected in gaps.

A. $3$ $3$
B. $2$ $2$
C. $4$ $4$
D. $1$ $1$

Solution

Related Formula

(R₁)/(R₂) = ( )/(100 - )

Core Logic

Initially, the meter bridge is balanced with $2 Ω$ $2\, \Omega$ and $3 Ω$ $3\, \Omega$:

(2)/(3) = (x)/(100 - x)

200 - 2x = 3x 5x = 200 x = 40 cm

Step 1: Shift after parallel connection

When an unknown resistor $R$ $R$ is connected in parallel with $3 Ω$ $3\, \Omega$, the new resistance in the right gap is $R' = (3R)/(3 + R)$ $R' = \frac{3R}{3 + R}$. The null point shifts by $22.5 cm$ $22.5 \text{ cm}$ towards Y, so the new balancing length is $x' = 40 + 22.5 = 62.5 cm$ $x' = 40 + 22.5 = 62.5 \text{ cm}$. The new balance equation is:

(2)/(R') = (62.5)/(100 - 62.5)

(2)/((3R)/(3 + R)) = (62.5)/(37.5) = (5)/(3)

(2(3 + R))/(3R) = (5)/(3)

$6 + 2R = 5R$ $6 + 2R = 5R$

3R = 6 R = 2 Ω

Pattern Recognition

A shift 'towards Y' means the balance point moved right, which implies the right gap resistance decreased. Connecting in parallel correctly decreases resistance.

Chapter Mix

Class 12 Physics: Current Electricity

Q26 jee_main_2026_24_january_evening DC Circuits and Ammeters

The reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ____ A.

DC Circuits and Ammeters diagram for Q26 - JEE Main 2026 Evening — A circuit diagram containing a 10V battery, multiple resistors, a 10 microfarad capacitor, and an ammeter.

A. $2$ $2$
B. $1$ $1$
C. $1 / 2$ $1 / 2$
D. $0$ $0$

Solution

Core Logic

In steady state, the capacitor acts as an open circuit. No current flows through the branch containing the $10$ $10\mu\text{F}$ capacitor.

After removing the capacitor branch, we simplify the network of resistors.

Step 1: Equivalent Resistance

I = 2A

By further simplifying the parallel and series combinations, we determine the main current and the current branching into the ammeter.

Step 2: Ammeter Reading

The current of $2A$ $2\mathrm{A}$ splits equally through the symmetrical $8Ω$ $8\Omega$ branches. The ammeter reading is $1A$ $1\mathrm{A}$.

Pattern Recognition

Steady state DC + Capacitor = Open circuit. Always delete the capacitor branch first before calculating equivalent resistance.

Chapter Mix

Class 12 Physics: Current Electricity Class 12 Physics: Electrostatics

Q29 jee_main_2026_24_january_evening Resistor Networks and Symmetry

A regular hexagon is formed by six wires each of resistance r

Ω

and the corners are joined to centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

A. $(4)/(5)r$ $\frac{4}{5}r$
B. $(5)/(8)r$ $\frac{5}{8}r$
C. $(3)/(4)r$ $\frac{3}{4}r$
D. $(3)/(5)r$ $\frac{3}{5}r$

Solution

Core Logic

By applying the principle of symmetry to the regular hexagonal resistor network, the circuit can be simplified across the axis of symmetry.

Resistor Networks and Symmetry diagram for Q29 - JEE Main 2026 Evening

Step 1: Network Reduction

Using equipotential nodes along the perpendicular bisector, we can resolve the network into parallel and series components.

The equivalent resistance for each half is determined as follows:

(R \times (R)/(2))/((3R)/(2)) = (R)/(3)

Step 2: Final Equivalent Resistance

Combining the simplified branches:

Req = (2R \times (4R)/(3))/(2R + (4R)/(3)) = (8R²)/(10R) = (4)/(5)R

Pattern Recognition

For regular polygonal networks with central connections, always search for a line of equipotential symmetry perpendicular to the line joining the input and output nodes. Nodes on this axis can be detached or merged.

Chapter Mix

Class 12 Physics: Current Electricity

Q48 jee_main_2026_24_january_evening Meter Bridge

In a meter bridge experiment to determine the value of unknown resistance, first the resistances

2 Ω

and

3 Ω

are connected in the left and right gaps of the bridge and the null point is obtained at a distance l cm from the left. Now when an unknown resistance x

Ω

is connected in parallel to

3 Ω

resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance x is ____

Ω

.

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

Meter bridge balance condition:

(R₁)/(R₂) = (l)/(100-l)

Core Logic

In Case I:

(2)/(3) = ( )/((100 - )) (1)

Step 1: Calculate initial null point

Solving Equation (1):

200 - 2 = 3

5 = 200 = 40 cm

Step 2: Apply Case II conditions

In Case II, the right gap resistance becomes a parallel combination: $R' = (3x)/(3+x)$ $R' = \frac{3x}{3+x}$. The new null point is shifted $10 cm$ $10 \text{ cm}$ to the right, meaning the new balancing length is $' = 40 + 10 = 50 cm$ $\ell' = 40 + 10 = 50 \text{ cm}$.

(2)/(R') = (50)/(100 - 50) = (50)/(50) = 1

Thus, $R' = 2 Ω$ $R' = 2 \, \Omega$.

Step 3: Solve for x

(3x)/(3 + x) = 2

$3x = 6 + 2x$ $3x = 6 + 2x$ $x = 6 Ω$ $x = 6 \, \Omega$

Pattern Recognition

If the null point falls exactly at $50 cm$ $50 \text{ cm}$, the two gap resistances must be identical. Recognizing this immediately yields $Rparallel = 2$ $R_{\text{parallel}} = 2$ without solving proportional fractions.

Chapter Mix

Class 12 Physics: Current Electricity

Q28 jee_main_2026_28_january_morning Potentiometer

In the potentiometer, when the cell in the secondary circuit is shunted with

4 Ω

resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell in shunted with

12 Ω

resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ____

Ω

.

A. $3$ $3$
B. $4$ $4$
C. $12$ $12$
D. $6$ $6$

Solution

Related Formula

For a cell of emf $E$ $E$ and internal resistance $r$ $r$ shunted by a resistance $R$ $R$, the terminal voltage is:

V = (E \cdot R)/(r + R)

In a potentiometer, the terminal voltage is proportional to the balancing length $l$ $l$: $V = K \cdot l$ $V = K \cdot l$

Core Logic

Equate the expressions for terminal voltage for both cases to their respective balancing lengths. Let $E$ $E$ be the emf and $r$ $r$ be the internal resistance of the cell.

Step 1: First Condition Setup

When shunted with $4 Ω$ $4 \,\Omega$:

(E \cdot 4)/(r + 4) = 120K --- (i)

Step 2: Second Condition Setup

When shunted with $12 Ω$ $12 \,\Omega$:

(E \cdot 12)/(r + 12) = 180K --- (ii)

Step 3: Solving for Internal Resistance

Divide equation (i) by equation (ii):

((4E)/(r + 4))/((12E)/(r + 12)) = (120K)/(180K)

(1)/(3) \cdot (r + 12)/(r + 4) = (2)/(3)

r + 12 = 2(r + 4)

$r + 12 = 2r + 8$ $r + 12 = 2r + 8$ $r = 4 Ω$ $r = 4 \,\Omega$

Pattern Recognition

Standard potentiometer problem. Ratio of balancing lengths maps directly to the ratio of terminal voltages. $(l₁)/(l₂) = (R₁(r+R₂))/(R₂(r+R₁))$ . Plugging the values simplifies it instantly.

Chapter Mix

Class 12 Physics: Current Electricity

A wire of resistance R$R$ 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

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Ratio Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Current Electricity Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Shift after parallel connection

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Equivalent Resistance

Step 2: Ammeter Reading

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Network Reduction

Step 2: Final Equivalent Resistance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate initial null point

Step 2: Apply Case II conditions

Step 3: Solve for x

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: First Condition Setup

Step 2: Second Condition Setup

Step 3: Solving for Internal Resistance

Pattern Recognition

Chapter Mix

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A wire of resistance $R$ $R$ 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