A regular hexagon is formed by six wires each of resistance r Omega and the corners are joined to centre by wires of…

Year	2026	2025	2024	Total
Questions	18	13	19	50

Q jee_main_2024_27_jan_morning Capacitor in DC Circuit

The charge accumulated on the capacitor connected in the following circuit is ______

. (Given

C = 150

).

Capacitor in DC Circuit diagram for Q59 - JEE Main 2024 Morning — The layout shows an integrated network of resistors labeled R1 to R4 powered by a 10V DC source loop with a branch housing a 150 uF capacitor element bridging distinct structural potential nodes.

Numerical Answer. Answer: 400 to 400

Solution

Related Formula

Q = C \cdot Δ VC

Core Logic

In steady state, no current flows through the capacitor branch. Analyze potential levels at node loops using standard Kirchhoff mesh values mapped across the core loop resistors:

VA + (10)/(3)(1) - 6(1) = VB

Step 1: Compute node potential difference

VA - VB = 6 - (10)/(3) = (8)/(3) V

Step 2: Evaluate accumulated charge

Q = C(VA - VB) = 150 \times (8)/(3) V = 50 \times 8 = 400

Pattern Recognition

Steady state capacitor loops act as open electrical cuts, transforming active mesh equations into simple layout node checks.

Chapter Mix

Class 12 Physics: Current Electricity

Q48 jee_main_2024_27_jan_morning Meter Bridge and Resistivity

A wire of length

10 cm

and radius

\sqrt(7) \times 10⁻⁴ m

is connected across the right gap of a meter bridge. When a resistance of

4.5 Ω

is connected on the left gap by using a resistance box, the balance length is found to be at

60 cm

from the left end. If the resistivity of the wire is

R \times 10⁻⁷

, then the value of

R

is:

A. $63$ $63$
B. $70$ $70$
C. $66$ $66$
D. $35$ $35$

Solution

Related Formula

From the balance condition of the meter bridge:

Xleftl = Xright100 - l

Resistance formula:

X = ρ (lw)/(A) = (ρ lw)/(π r²)

Core Logic

First, evaluate the unknown resistance $Xright$ $X_{\text{right}}$ in the right gap:

(4.5)/(60) = Xright40 Xright = (4.5 \times 40)/(60) = 3 Ω

Step 1: Calculate Resistivity Value

Now map the resistance parameters ( $lw = 10 cm = 0.1 m$ $l_w = 10\text{ cm} = 0.1\text{ m}$, $r = \sqrt(7) \times 10⁻⁴ m$ $r = \sqrt{7} \times 10^{-4}\text{ m}$):

3 = ρ 0.1(22)/(7) \times (\sqrt(7) \times 10⁻⁴)²

3 = ρ 0.1(22)/(7) \times 7 \times 10⁻⁸

3 = ρ 0.122 \times 10⁻⁸

ρ = 3 \times 22 \times 10⁻⁸0.1 = 66 \times 10⁻⁷

Step 2: Compare to find R

Given $ρ = R \times 10⁻⁷$ $\rho = R \times 10^{-7}$, comparing coefficients gives:

$R = 66$ $R = 66$

Pattern Recognition

Meter bridge balance simplifies directly to simple scalar component checks. The $\sqrt(7)$ $\sqrt{7}$ term perfectly neutralizes the fractional $(22)/(7)$ $\frac{22}{7}$ constant in circular area profiles.

Chapter Mix

Class 12 Physics: Current Electricity

Q49 jee_main_2024_27_jan_morning Combination of Resistors

A wire of resistance

R

and length

L

is cut into

5

equal parts. If these parts are joined parallely, then the resultant resistance will be:

A. $(1)/(25)R$ $\frac{1}{25}R$
B. $(1)/(5)R$ $\frac{1}{5}R$
C. $25R$ $25R$
D. $5R$ $5R$

Solution

Core Logic

Resistance is directly proportional to length ( $R \propto L$ $R \propto L$). Cutting the wire into 5 equal pieces reduces the resistance of each segment to:

R' = (R)/(5)

Step 1: Compute parallel value

Connecting 5 identical resistors $R'$ $R'$ in parallel gives a total equivalent resistance of:

Req = (R')/(5) = (R/5)/(5) = (R)/(25)

Pattern Recognition

Cutting an item into $N$ $N$ components and grouping them in parallel scales the overall baseline systemic value down cleanly by a factor of $N²$ $N^2$.

Chapter Mix

Class 12 Physics: Current Electricity

Q jee_main_2024_29_jan_morning Kirchhoff's Laws and RC Circuits

A

16 Ω

wire is bend to form a square loop. A 9V battery with internal resistance

1 Ω

is connected across one of its sides. If a

4 μ F

capacitor is connected across one of its diagonals, the energy stored by the capacitor will be

(x)/(2) μ J

. where

x =

_________.

Numerical Answer. Answer: 81 to 81

Solution

Related Formula

Energy stored ( $U$ $U$) by a capacitor of capacitance $C$ $C$ under steady state voltage $VC$ $V_C$:

U = (1)/(2) C VC²

Core Logic

A wire of resistance $16 Ω$ $16 \, \Omega$ is bent into a square, so each of the 4 sides has a resistance of:

Rside = (16)/(4) = 4 Ω

The battery is connected across one side. Let this side have resistance $4 Ω$ $4 \, \Omega$. The remaining three sides are connected in series, creating a parallel branch with a combined resistance of:

Rₛₑᵣᵢₑₛ = 4 + 4 + 4 = 12 Ω

Circuit network layout showing the square loop resistance divisions and diagonal capacitor connection for Q57

Step 1: Calculate Equivalent Resistance and Circuit Current

The equivalent external resistance of the parallel loop branches is:

Rₚ = (12 \times 4)/(12 + 4) = (48)/(16) = 3 Ω

Including the battery's internal resistance ( $1 Ω$ $1 \, \Omega$), the total line current $I$ $I$ leaving the battery is:

I = (V)/(Rₚ + r) = (9)/(3 + 1) = (9)/(4) ~A

Step 2: Find Current through the Main Branches

Using the current divider rule, the current $I₁$ $I_1$ flowing through the longer $12 Ω$ $12 \, \Omega$ branch is:

I₁ = I \times (4)/(12 + 4) = (9)/(4) \times (4)/(16) = (9)/(16) ~A

Step 3: Find Potential Difference across the Diagonal

In a steady state, the capacitor acts as an open circuit. Let the diagonal link nodes be $A$ $A$ and $B$ $B$. The path contains two sides of the $12 Ω$ $12 \, \Omega$ branch (total resistance = $8 Ω$ $8 \, \Omega$):

VC = VA - VB = I₁ \times 8 = (9)/(16) \times 8 = (9)/(2) ~V

Step 4: Compute Stored Energy and Extract x

The stored energy $U$ $U$ is:

U = (1)/(2) \times (4 μ F) \times ((9)/(2))² = (1)/(2) \times 4 \times (81)/(4) = (81)/(2) μ J

Comparing this directly with the expression $(x)/(2) μ J$ $\frac{x}{2} \, \mu \mathrm{J}$:

$x = 81$ $x = 81$

Therefore, the value of $x$ $x$ is $81$ $81$.

Pattern Recognition

Remember that capacitors act as standard open circuits when reaching steady-state DC conditions. Simply calculate the node potentials across the bridge connection points using standard current distribution laws first, then execute the energy equation.

Chapter Mix

Class 12 Physics: Current Electricity Class 12 Physics: Electrostatic Potential and Capacitance

Q32 jee_main_2024_29_jan_morning Electric Current and Charge

The electric current through a wire varies with time as

I = I₀ + β t

, where

I₀ = 20 ~A

and

β = 3 ~A / s

. The amount of electric charge crossed through a section of the wire in

20 ~s

is:

A. $80 ~C$ $80 \mathrm{~C}$
B. $1000 ~C$ $1000 \mathrm{~C}$
C. $800 ~C$ $800 \mathrm{~C}$
D. $1600 ~C$ $1600 \mathrm{~C}$

Solution

Related Formula

The relationship between current, charge, and time is given by:

I = (dq)/(dt) q = \int I dt

Core Logic

Given the current variation:

I = I₀ + β t

Substituting the given values, $I₀ = 20 ~A$ $I_0 = 20 \mathrm{~A}$ and $β = 3 ~A/s$ $\beta = 3 \mathrm{~A/s}$:

$I = 20 + 3t$ $I = 20 + 3t$

Thus:

(dq)/(dt) = 20 + 3t

Step 1: Integrate to Find Charge

To find the total charge crossing a section of wire from $t = 0$ $t = 0$ to $t = 20 ~s$ $t = 20 \mathrm{~s}$:

\int₀q dq = \int₀²⁰ (20 + 3t) dt

q = [ 20t + (3t²)/(2) ]₀²⁰

q = 20(20) + (3(20)²)/(2)

q = 400 + (3(400))/(2) = 400 + 600 = 1000 ~C

Therefore, the charge crossed is $1000 ~C$ $1000 \mathrm{~C}$.

Pattern Recognition

Whenever current is given as a function of time $I(t)$ $I(t)$, the total charge is simply the area under the $I-t$ $I-t$ curve, which mathematically corresponds to the definite integral $\intt₁t₂ I(t) dt$ $\int_{t_1}^{t_2} I(t) \, dt$.

Chapter Mix

Class 12 Physics: Current Electricity

A regular hexagon is formed by six wires each of resistance r Ω$\Omega$ 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

Step 1: Network Reduction

Step 2: Final Equivalent Resistance

Pattern Recognition

Chapter Mix

Reference Study Guides

More Current Electricity Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Compute node potential difference

Step 2: Evaluate accumulated charge

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Resistivity Value

Step 2: Compare to find R

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Compute parallel value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Equivalent Resistance and Circuit Current

Step 2: Find Current through the Main Branches

Step 3: Find Potential Difference across the Diagonal

Step 4: Compute Stored Energy and Extract x

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integrate to Find Charge

Pattern Recognition

Chapter Mix

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A regular hexagon is formed by six wires each of resistance r $Ω$ $\Omega$ 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