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

Q25 jee_main_2025_03_april_morning Galvanometer/Ammeter with Shunt Resistance

In the figure shown below, a resistance of

150.4Ω

is connected in series to an ammeter A of resistance

240Ω

. A shunt resistance of

10Ω

is connected in parallel with the ammeter. The reading of the ammeter is ________

mA

.

Series parallel circuit showing ammeter with shunt for Q25 — A circuit schematic containing a 20V battery connected in series with a 150.4 ohm resistor and a parallel combination of a 240 ohm ammeter and a 10 ohm shunt resistor.

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

Parallel combination of ammeter ( $RG = 240Ω$ $R_G = 240\Omega$) and shunt ( $S = 10Ω$ $S = 10\Omega$):

Rparallel = (RG \cdot S)/(RG + S)

Total equivalent resistance:

Req = Rₛₑᵣᵢₑₛ + Rparallel

Current Divider Rule:

Iammeter = Itotal \cdot ((S)/(RG + S))

Core Logic

Let's first calculate the equivalent resistance of the parallel branch (ammeter + shunt):

Rparallel = (240 \times 10)/(240 + 10) = (2400)/(250) = 9.6Ω

Now, add the series resistance ( $150.4Ω$ $150.4\Omega$) to find the total equivalent resistance of the circuit:

Req = 150.4 + 9.6 = 160Ω

Step 1: Finding Total Current

Use Ohm's Law to calculate the total current $Itotal$ $I_{\text{total}}$ drawn from the $20~V$ $20\mathrm{~V}$ battery:

Itotal = VReq = (20)/(160) = 0.125~A

Step 2: Calculating Ammeter Current

Using the Current Divider Rule, calculate the current $Iammeter$ $I_{\text{ammeter}}$ flowing through the $240Ω$ $240\Omega$ ammeter branch:

Iammeter = Itotal \times (10)/(240 + 10) = 0.125 \times (10)/(250)

Iammeter = 0.125 \times (1)/(25) = 0.005~A

Convert this into milliamperes ( $mA$ $\mathrm{mA}$):

Iammeter = 0.005 \times 1000 = 5~mA

Therefore, the reading of the ammeter is $5~mA$ $5\mathrm{~mA}$.

Simplified equivalent circuit showing ammeter shunt current split for Q25 — A circuit schematic containing a 20V battery connected in series with a 150.4 ohm resistor and a parallel combination of a 240 ohm ammeter and a 10 ohm shunt resistor.

Pattern Recognition

Notice how nicely the numbers are set up in JEE Mains questions! The parallel combination of $240Ω$ $240\Omega$ and $10Ω$ $10\Omega$ yields exactly $9.6Ω$ $9.6\Omega$, which beautifully combines with $150.4Ω$ $150.4\Omega$ to form a perfect integer sum of $160Ω$ $160\Omega$. This tells you that your intermediate steps are absolutely correct!

Chapter Mix

Class 12 Physics: Current Electricity: Measuring Devices

Q2 jee_main_2025_04_april_evening Electric Power

There are 'n' number of identical electric bulbs, each is designed to draw a power p independently from the mains supply. They are now joined in series across the main supply. The total power drawn by the combination is:

A. $np$ $np$
B. $pn²$ $\frac{p}{n^{2}}$
C. $(p)/(n)$ $\frac{p}{n}$
D. $p$ $p$

Solution

Related Formula

P = (V²)/(R)

For series combination:

Rₛ = R₁ + R₂ + + Rₙ

Core Logic

Since the bulbs are identical, each has a resistance $R = (V²)/(p)$ $R = \frac{V^2}{p}$. When $n$ $n$ identical bulbs are connected in series, the total equivalent resistance becomes: $Rₛ = nR$ $R_s = nR$

Step 1: Calculate Combined Power

The total power drawn across the same mains supply $V$ $V$ is:

Pₛ = (V²)/(Rₛ) = (V²)/(nR) = (1)/(n) ((V²)/(R)) = (p)/(n)

Pattern Recognition

Identical appliances connected in series scale down their combined power inversely with count ( $Pₙₑₜ = P/n$ $P_{net} = P/n$), identical appliances in parallel scale up combined power linearly ( $Pₙₑₜ = nP$ $P_{net} = nP$).

Chapter Mix

Class 12 Physics: Current Electricity

Q6 jee_main_2025_04_april_evening Combination of Resistors

From the combination of resistors with resistance values

R₁=R₂=R₃=5 Ω

and

R₄=10 Ω

, which of the following combination is the best circuit to get an equivalent resistance of

6 Ω

?

A.
B.
C.
D.

Solution

Related Formula

Series combination: $Rₛ = Rₐ + Rb$ $R_s = R_a + R_b$ Parallel combination:

(1)/(Rₚ) = (1)/(R₁) + (1)/(R₂) Rₚ = (R₁ R₂)/(R₁ + R₂)

Core Logic

Let's check option (1): Top branch has $R₁$ $R_1$ and $R₂$ $R_2$ in series: $Rₜₒₚ = 5 + 5 = 10 Ω$ $R_{\text{top}} = 5 + 5 = 10\ \Omega$. Bottom branch has $R₃$ $R_3$ and $R₄$ $R_4$ in series: $Rbottom = 5 + 10 = 15 Ω$ .

Step 1: Calculate Equivalent Parallel Resistance

These two branches are connected in parallel, so the total equivalent resistance is:

Rₚ = (10 \times 15)/(10 + 15) = (150)/(25) = 6 Ω

This matches the target value of $6 Ω$ $6\ \Omega$ perfectly.

Circuit diagram analysis for 6 ohm equivalent resistance

Pattern Recognition

Look for symmetric partitions. Standard combinations of values like $10 Ω$ $10\ \Omega$ and $15 Ω$ $15\ \Omega$ yield exactly $6 Ω$ $6\ \Omega$ in parallel.

Chapter Mix

Class 12 Physics: Current Electricity

Q jee_main_2025_04_april_morning Electric Current and Charge Flow

Current passing through a wire as function of time is given as

I(t)=0.02t+0.01~A

. The charge that will flow through the wire from

t=1~s

to

t=2~s

is:

A. 0.06 C
B. 0.02 C
C. 0.07 C
D. 0.04 C

Solution

Related Formula

q = \intt₁t₂ I(t) dt

Core Logic

Given: $I(t) = (0.02t + 0.01) A$ $I(t) = (0.02t + 0.01)\text{ A}$ Limits: $t₁ = 1 s$ $t_1 = 1\text{ s}$, $t₂ = 2 s$ $t_2 = 2\text{ s}$

Step 1: Perform Definite Integration

q = \int₁² (0.02t + 0.01) dt

q = [ 0.02(t²)/(2) + 0.01t ]₁² = [ 0.01t² + 0.01t ]₁²

q = [ 0.01(2)² + 0.01(2) ] - [ 0.01(1)² + 0.01(1) ]

q = [0.04 + 0.02] - [0.01 + 0.01] = 0.06 - 0.02 = 0.04 C

Hence, the total charge flowing through the wire is $0.04 C$ $0.04\text{ C}$.

Pattern Recognition

Definite integration of a linear current function can also be verified geometrically by calculating the area of the trapezoid under the $I--t$ $I\text{--}t$ curve:

Area = (I(1) + I(2))/(2) \times (2 - 1) = (0.03 + 0.05)/(2) \times 1 = 0.04 C

Evaluation Rubric / Model Answer

Option D: 0.04 C

Chapter Mix

Class 12 Physics: Current Electricity

Q23 jee_main_2025_24_jan_morning Combination of Resistors

A wire of resistance

9Ω

is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ______ ohm.

Numerical Answer. Answer: 2 to 2

Solution

Core Logic

The total resistance of the continuous uniform wire loop is $9Ω$ $9\Omega$. When bent into an equilateral triangle, it is split into three equal length sections. The resistance of each individual side is:

Rside = (9Ω)/(3) = 3Ω

Step 1: Calculating Equivalent Series and Parallel Resistance

As shown in the circuit diagrams

Combination of Resistors diagram for Q23 - JEE Main 2025 Morning

and

, measuring across any two vertices means one branch contains a single side resistor (

3Ω

), while the other branch contains the remaining two sides connected in series :

Rₛₑᵣᵢₑₛ = 3Ω + 3Ω = 6Ω

Now, calculate the parallel equivalent between these two branches:

Req = Rside \times RₛₑᵣᵢₑₛRside + Rₛₑᵣᵢₑₛ = (3 \times 6)/(3 + 6) = (18)/(9) = 2Ω

Pattern Recognition

For a closed uniform loop with $N$ $N$ equal sides, the parallel resistance measured across adjacent corners always simplifies to $(N-1)/(N²) \cdot Rtotal$ .

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 6

Solution

Related Formula

Core Logic

Step 1: Finding Total Current

Step 2: Calculating Ammeter Current

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Combined Power

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Equivalent Parallel Resistance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Perform Definite Integration

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Core Logic

Step 1: Calculating Equivalent Series and Parallel 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