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

Q jee_main_2024_30_january_evening Voltmeter Resistance

Two resistance of

100Ω

and

200Ω

are connected in series with a battery of

4 ~V

and negligible internal resistance. A voltmeter is used to measure voltage across

100Ω

resistance, which gives reading as

1 ~V

. The resistance of voltmeter must be ________

Ω

.

Numerical Answer. Answer: 200 to 200

Solution

Related Formula

V = I Req

Rparallel = (R₁ R₂)/(R₁ + R₂)

Core Logic

Voltmeter Resistance diagram for Q60 - JEE Main 2024 Evening

The voltmeter has some internal resistance $Rv$ $R_v$ and is connected in parallel with the $100Ω$ $100\Omega$ resistor. The equivalent resistance of this parallel combination is $Rₚ = (100 Rv)/(100 + Rv)$ $R_p = \frac{100 R_v}{100 + R_v}$. This combination is in series with the $200Ω$ $200\Omega$ resistor. The total voltage applied is $4 ~V$ $4 \mathrm{~V}$.

Step 1: Set up Voltage Divider

The voltage across the parallel combination (the voltmeter reading) is $1 ~V$ $1 \mathrm{~V}$. Therefore, the voltage across the $200Ω$ $200\Omega$ resistor must be $4 ~V - 1 ~V = 3 ~V$ . Using the voltage divider rule (or equating currents since they are in series):

I = V₂₀₀200 = (3)/(200) ~A

Step 2: Solve for Voltmeter Resistance

The current $I$ $I$ also flows through the parallel combination $Rₚ$ $R_p$: $Vₚ = I Rₚ$ $V_p = I R_p$

1 = ((3)/(200)) ( (100 Rv)/(100 + Rv) )

200 (100 + Rv) = 300 Rv

20000 + 200 Rv = 300 Rv

$100 Rv = 20000$ $100 R_v = 20000$

Rv = 200 Ω

Pattern Recognition

If the voltage splits as $1V$ $1\mathrm{V}$ to $3V$ $3\mathrm{V}$, the resistances must be in a $1:3$ $1:3$ ratio. So $Rₚ = 200 / 3$ $R_p = 200 / 3$. Equating $100 Rv / (100+Rv) = 200/3$ $100 R_v / (100+R_v) = 200/3$ instantly gives $Rv = 200$ $R_v = 200$.

Chapter Mix

Class 12 Physics: Current Electricity

Q48 jee_main_2024_30_january_evening Power Dissipation in Resistors

When a potential difference

V

is applied across a wire of resistance

R

, it dissipates energy at a rate

W

. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:

A. $1 / 4 W$ $1 / 4 \mathrm{W}$
B. $1 / 2 W$ $1 / 2 \mathrm{W}$
C. $2 ~W$ $2 \mathrm{~W}$
D. $4 ~W$ $4 \mathrm{~W}$

Solution

Related Formula

P = V²Req

Core Logic

Initially, the power (rate of energy dissipation) is $W = (V²)/(R)$ $W = \frac{V^2}{R}$. When the wire is cut into two halves, each half will have a resistance of $(R)/(2)$ $\frac{R}{2}$ (since resistance is directly proportional to length).

Step 1: Equivalent Resistance

When these two halves (each of $(R)/(2)$ $\frac{R}{2}$) are connected in parallel, their equivalent resistance $Req$ $R_{\text{eq}}$ is:

Req = (((R)/(2)) \times ((R)/(2)))/((R)/(2) + (R)/(2)) = (R)/(4)

Step 2: New Power Dissipation

The new rate of energy dissipation $W'$ $W'$ across the same potential difference $V$ $V$ is:

W' = V²Req = (V²)/((R)/(4)) = 4 (V²)/(R)

$W' = 4W$ $W' = 4W$

Pattern Recognition

Cutting a resistor into $n$ $n$ equal parts and connecting them in parallel reduces the total resistance by a factor of $n²$ $n^2$. Consequently, for a constant voltage supply, the power dissipated increases by a factor of $n²$ $n^2$.

Chapter Mix

Class 12 Physics: Current Electricity

Q jee_main_2024_30_jan_morning Cells in Opposition and Terminal Voltage

Two cells are connected in opposition as shown. Cell

E₁

is of

8V

emf and

2Ω

internal resistance; the cell

E₂

is of

2V

emf and

4Ω

internal resistance. The terminal potential difference of cell

E₂

is:

Cells in Opposition and Terminal Voltage diagram for Q52 - JEE Main 2024 Morning — Two batteries pushing current against each other.

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

I = EₙₑₜReq

V = E - Ir (Discharging)

V = E + Ir (Charging)

Core Logic

Circuit with marked nodes for Kirchhoff analysis. — Two batteries pushing current against each other.

Because the cells are in opposition, the net EMF drives current from the higher potential cell (

8V

) to the lower potential cell (

2V

). Thus, the

2V

cell acts as a load and undergoes charging.

Step 1: Calculate Total Current

I = (8 - 2)/(2 + 4) = (6)/(6) = 1 ~A

Step 2: Terminal Potential of Cell 2

Since cell $E₂$ $E_2$ is being charged, its terminal potential difference is:

V₂ = E₂ + I r₂

Applying Kirchhoff's rule across cell $E₂$ $E_2$ (from node C to B):

VC - VB = E₂ + I r₂ = 2 + (1)(4) = 6 ~V

Pattern Recognition

When a smaller battery is forced backwards by a larger battery, it gets "charged". Consequently, its terminal voltage increases above its nominal EMF: $V = E + Ir$ $V = E + Ir$.

Chapter Mix

Class 12 Physics: Current Electricity

Q33 jee_main_2024_30_jan_morning Resistors in Series and Voltage Dividers

A potential divider circuit is shown in figure. The output voltage

V₀

is

Resistors in Series and Voltage Dividers diagram for Q33 - JEE Main 2024 Morning — Circuit containing multiple resistors in series and parallel calculating a specific output voltage.

A. $4 ~V$ $4 \mathrm{~V}$
B. $2 ~mV$ $2 \mathrm{~mV}$
C. $0.5 ~V$ $0.5 \mathrm{~V}$
D. $12 ~mV$ $12 \mathrm{~mV}$

Solution

Related Formula

$V = IR$ $V = IR$

Req = R₁ + R₂ + + Rₙ (for series)

Core Logic

Observe the circuit diagram. The total equivalent resistance $Req$ $R_{\text{eq}}$ of the series network must be calculated by summing all the resistance values shown in the main branch.

Step 1: Calculate Total Resistance and Current

From the given network, assuming the total series resistance is $4000 Ω$ $4000 \,\Omega$ (comprising the $3.3kΩ$ $3.3\mathrm{k}\Omega$ resistor and seven $100 Ω$ $100\,\Omega$ resistors).

Req = 4000 Ω

The total voltage applied across the network is $4 ~V$ $4 \mathrm{~V}$.

i = VReq = (4)/(4000) = (1)/(1000) ~A

Step 2: Calculate Output Voltage

The output voltage $V₀$ $V_0$ is tapped across five $100 Ω$ $100 \,\Omega$ resistors.

Rout = 5 \times 100 Ω = 500 Ω

Thus, the output voltage is:

V₀ = i \cdot Rout = ((1)/(1000)) \times 500 = 0.5 ~V

Pattern Recognition

A potential divider simply scales the input voltage by the fraction of the resistance tapped over the total resistance: $V₀ = Vᵢₙ \times (Rₜₐₚ / Rtotal)$ . Standard DC circuit division.

Chapter Mix

Class 12 Physics: Current Electricity

Q43 jee_main_2024_30_jan_morning Temperature Dependence of Resistivity

An electric toaster has resistance of

60Ω

at room temperature

(27°C)

. The toaster is connected to a

220V

supply. If the current flowing through it reaches

2.75A

, the temperature attained by toaster is around: (if

α = 2\times 10⁻⁴ / °C

)

A. $694° C$ $694^{\circ} \mathrm{C}$
B. $1235°C$ $1235^{\circ}\mathrm{C}$
C. $1694°C$ $1694^{\circ}\mathrm{C}$
D. $1667°C$ $1667^{\circ}\mathrm{C}$

Solution

Related Formula

R = (V)/(I)

R = R₀ (1 + α Δ T)

Core Logic

First, evaluate the final resistance $RT$ $R_T$ at the operating condition using Ohm's law. Second, plug the final resistance into the linear temperature dependence equation for resistance to solve for final temperature $T$ $T$.

Step 1: Calculate Final Resistance

Given $V = 220 ~V$ $V = 220 \mathrm{~V}$ and $I = 2.75 ~A$ $I = 2.75 \mathrm{~A}$:

RT = (220)/(2.75) = 80 Ω

Step 2: Apply Temperature Equation

We know $R₂₇ = 60 Ω$ $R_{27} = 60 \,\Omega$.

RT = R₂₇ [1 + α (T - 27)]

80 = 60 [1 + 2 \times 10⁻⁴ (T - 27)]

(80)/(60) = 1 + 2 \times 10⁻⁴ (T - 27)

(4)/(3) - 1 = 2 \times 10⁻⁴ (T - 27)

(1)/(3) = 2 \times 10⁻⁴ (T - 27)

Step 3: Solve for T

T - 27 = 16 \times 10⁻⁴

T - 27 = (10000)/(6) = 1666.67

T = 1666.67 + 27 \approx 1693.67 \Rightarrow 1694^° C

Pattern Recognition

Always separate the final temperature $T$ $T$ from $Δ T$ $\Delta T$. The most common error is forgetting to add back the initial reference temperature ( $27^$ $27^\circ\mathrm{C}$) at the end.

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 9

Solution

Related Formula

Core Logic

Step 1: Set up Voltage Divider

Step 2: Solve for Voltmeter Resistance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equivalent Resistance

Step 2: New Power Dissipation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Total Current

Step 2: Terminal Potential of Cell 2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Total Resistance and Current

Step 2: Calculate Output Voltage

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Final Resistance

Step 2: Apply Temperature Equation

Step 3: Solve for T

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