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Current Electricity appeared 50 times across 3 years — 5.8% of Physics. This question is from Resistor Networks and Symmetry.

Year 2026 2025 2024 Total
Questions 18 13 19 50

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

Solution & Explanation

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

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

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

The equivalent resistance for each half is determined as follows:

(R × (R)/(2))/((3R)/(2)) = (R)/(3)
Step 2: Final Equivalent Resistance

Combining the simplified branches:

Req = (2R × (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

Reference Study Guides

More Current Electricity Previous-Year Questions — Page 10

Q39 jee_main_2024_31_jan_evening Meter Bridge
The resistance per centimeter of a meter bridge wire is r, with X Ω resistance in left gap. Balancing length from left end is at 40 cm with 25 Ω resistance in right gap. Now the wire is replaced by another wire of 2r resistance per centimeter. The new balancing length for same settings will be at
  • A. 20 cm
  • B. 10 cm
  • C. 80 cm
  • D. 40 cm

Solution

Related Formula
RleftRwire-left = RrightRwire-right
Core Logic

For a meter bridge, the balancing condition is independent of the absolute resistance of the bridge wire as long as it is uniform. The ratio of the resistances in the gaps balances with the ratio of lengths.

Meter Bridge diagram for Q39 - JEE Main 2024 Evening
Meter Bridge diagram for Q39 - JEE Main 2024 Evening

Step 1: First Condition
(X)/(r ₁) = (25)/(r (100 - ₁))

Given ₁ = 40 cm:

(X)/(r × 40) = (25)/(r × 60) (X)/(40) = (25)/(60)
Step 2: Second Condition

When replaced by a wire of 2r per cm, the new lengths ₂ will satisfy:

(X)/(2r ₂) = (25)/(2r (100 - ₂))

Notice that the 2r terms cancel out entirely from both sides, leaving:

(X)/( ₂) = (25)/(100 - ₂)
Step 3: Conclusion

Since the ratio X/25 remains identical, the balancing length ratio / (100- ) also remains identical. Therefore, ₂ = ₁ = 40 cm.

Pattern Recognition

Meter bridge balance point strictly depends on length ratio, NOT the specific resistivity or thickness of the wire (provided it is uniform). If external resistors don't change, the balance point never changes.

Chapter Mix

Class 12 Physics: Current Electricity

Q43 jee_main_2024_31_jan_evening Power Dissipation
By what percentage will the illumination of the lamp decrease if the current drops by 20%?
  • A. 46%
  • B. 26%
  • C. 36%
  • D. 56%

Solution

Related Formula

Power (Illumination) is proportional to the square of the current for a constant resistance: P = I² R

Core Logic

Initial power: P₁ = I₁² R

If current drops by 20%, the new current is:

I₂ = I₁ - 0.2 I₁ = 0.8 I₁

New power:

P₂ = (0.8 I₁)² R = 0.64 I₁² R = 0.64 P₁
Step 1: Calculate Percentage Change
Δ P % = (P₂ - P₁)/(P₁) × 100% Δ P % = (0.64 P₁ - P₁)/(P₁) × 100% Δ P % = (0.64 - 1) × 100% = -36%
Step 2: Final Statement

The negative sign indicates a decrease. The illumination drops by 36%.

Pattern Recognition

For squared relations y = x², if x changes by a factor k (0.8), y changes by a factor k² (0.64). 1 - 0.64 = 36%. This bypasses algebraic limits usually done for small changes (like 2Δ x / x) since 20% is too large for approximation.

Chapter Mix

Class 12 Physics: Current Electricity

Q51 jee_main_2024_31_jan_evening Power in DC Circuits
In the following circuit, the battery has an emf of 2 V and an internal resistance of (2)/(3) Ω. The power consumption in the entire circuit is ______ W.
Power in DC Circuits diagram for Q51 - JEE Main 2024 Evening
The image shows a DC circuit with multiple resistors in parallel/series connected to a 2V battery with 2/3 ohm internal resistance.
Numerical Answer. Answer: 3 to 3

Solution

Related Formula
P = V²Req
Core Logic

To find total power, collapse the entire external circuit and battery internal resistance into a single equivalent resistance Req across the battery's ideal terminals.

Step 1: Equivalent Resistance Calculation

Analyzing the diagram: The circuit simplifies to an equivalent resistance Req combining the parallel/series elements along with the internal resistance r = 2/3 Ω. The final simplified equivalent resistance of the entire system calculates to:

Req = (4)/(3) Ω
Step 2: Calculate Power
P = V²Req P = (2²)/(4/3) P = (4)/(4/3) = 3 W
Pattern Recognition

Whenever "entire circuit" power is asked, include the battery's internal resistance inside Req so you can use P = E² / Rtotal directly.

Chapter Mix

Class 12 Physics: Current Electricity

Q jee_main_2024_31_jan_morning Resistor Circuits
Equivalent resistance of the following network is ________ Ω
Resistor Circuits diagram for Q53 - JEE Main 2024 Morning
A complex resistor bridge network connecting nodes A and B with multiple branches.
Numerical Answer. Answer: 1 to 1

Solution

Related Formula
Rparallel = (1)/(Σ (1)/(Rᵢ))
Core Logic

Resistor Circuits diagram for Q53 - JEE Main 2024 Morning
A complex resistor bridge network connecting nodes A and B with multiple branches.

By carefully identifying the nodes, we can see that a 6 Ω resistor in the middle branch is short-circuited by a direct zero-resistance wire path across it.

Resistor Circuits diagram for Q53 - JEE Main 2024 Morning
A complex resistor bridge network connecting nodes A and B with multiple branches.

Removing the short-circuited 6 Ω resistor simplifies the circuit into three identical branches connected directly between the terminals A and B.

Resistor Circuits diagram for Q53 - JEE Main 2024 Morning
A complex resistor bridge network connecting nodes A and B with multiple branches.

Step 2: Equivalent Calculation

The simplified circuit consists of three identical 3 Ω resistors in parallel.

Req = 3 × (1)/(3) = 1 Ω
Pattern Recognition

Always trace nodes directly connected by straight wires (zero resistance). Any resistor with both ends connecting to the exact same electrical node is shorted out and can be erased.

Chapter Mix

Class 12 Physics: Current Electricity

Q48 jee_main_2024_31_jan_morning Temperature Dependence Of Resistance
Two conductors have the same resistances at 0°C but their temperature coefficients of resistance are α₁ and α₂. The respective temperature coefficients for their series and parallel combinations are :
  • A. α₁ + α₂, (α₁ + α₂)/(2)
  • B. (α₁ + α₂)/(2), (α₁ + α₂)/(2)
  • C. α₁ + α₂, (α₁α₂)/(α₁ + α₂)
  • D. (α₁ + α₂)/(2), α₁ + α₂

Solution

Related Formula
RT = R₀(1 + α Δ T)
Step 1: Series Combination

Let base resistance at 0^° C be R. For series:

Req = R₁ + R₂ (2R)[1 + αeq,s Δ T] = R(1 + α₁ Δ T) + R(1 + α₂ Δ T) 2 + 2αeq,s Δ T = 2 + (α₁ + α₂)Δ T αeq,s = (α₁ + α₂)/(2)
Step 2: Parallel Combination

For parallel at 0^° C, Req,0 = R/2.

Req,p = (R₁ R₂)/(R₁ + R₂) (R)/(2) [1 + αeq,p Δ T] = (R² (1 + α₁ Δ T)(1 + α₂ Δ T))/(R(2 + (α₁ + α₂)Δ T)) (1)/(2) (1 + αeq,p Δ T) = (1 + (α₁ + α₂)Δ T)/(2(1 + (α₁ + α₂)/(2)Δ T))

Using binomial expansion for small Δ T:

1 + αeq,p Δ T ≈ [1 + (α₁ + α₂)Δ T] [ 1 - (α₁ + α₂)/(2)Δ T ] 1 + αeq,p Δ T ≈ 1 + (α₁ + α₂)Δ T - (α₁ + α₂)/(2)Δ T αeq,p Δ T = (α₁ + α₂)/(2)Δ T αeq,p = (α₁ + α₂)/(2)
Pattern Recognition

For two identical base resistances, the effective temperature coefficient is simply the arithmetic mean of their individual coefficients, regardless of whether they are in series or parallel.

Chapter Mix

Class 12 Physics: Current Electricity

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