A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R1 = R2 = R3 = R4 ). When R3 resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference ( Va - Vb ) (after R3 is heated) is ____ V.

Initially, R1 = R2 = R3 = R4 = R. The bridge is balanced, and VA = VB. After heating, R3 increases by 10%, meaning the new R3' = R + 0.1R = 1.1R. The other arms remain R. The battery provides a voltage of V = 40 V. Let the lower node (D) be at 0 V and the upper node (C) be at 40 V. The potential at node A (between R3' and R4) can be found via the voltage divider rule across the right branch (R3' and R4). Wait, examining the circuit labels carefully: Assume the branches are: Left branch contains R1 and R2. Right branch contains R3 and R4. Node B is between R1 and R2. Node A is between R3 and R4. From the provided solution: This implies the branch containing node A consists of two equal resistances R, which contradicts the textual convention if R3 changed. Looking at the provided solution image logic, the branch with B has the changed resistance 1.1R. Let's follow the PDF's mathematical layout exactly. (Meaning branch A has equal resistors R and R) *(Note: The PDF solution computes VB = V / 2.1, implying the resistance below node B is R, and above is 1.1R)* The closest match among the options is 0.95 V. For a slightly unbalanced Wheatstone bridge due to a fractional change Delta R/R = x, the potential difference Delta V is approximately V cdot (x/4) for small x. Here x=0.1, so Delta V approx 40 x 0.1 / 4 = 1 V. Exact calculation gives 0.95 V. Class 12 Physics: Current Electricity

A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R1 = R2 = R3 = R4 ). When R3 resistance is heated to some temperature, its resistance v

Initially, R1 = R2 = R3 = R4 = R. The bridge is balanced, and VA = VB. After heating, R3 increases by 10%, meaning the new R3' = R + 0.1R = 1.1R. The other arms remain R. The battery provides a voltage of V = 40 V. Let the lower node (D) be at 0 V and the upper node (C) be at 40 V. The potential at node A (between R3' and R4) can be found via the voltage divider rule across the right branch (R3' and R4). Wait, examining the circuit labels carefully: Assume the branches are: Left branch contains

A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R1 = R2 = R3 = R4 ). When R3 resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference ( Va - Vb ) (after R3 is heated) is ____ V.

Initially, R1 = R2 = R3 = R4 = R. The bridge is balanced, and VA = VB. After heating, R3 increases by 10%, meaning the new R3' = R + 0.1R = 1.1R. The other arms remain R. The battery provides a voltage of V = 40 V. Let the lower node (D) be at 0 V and the upper node (C) be at 40 V. The potential at node A (between R3' and R4) can be found via the voltage divider rule across the right branch (R3' and R4). Wait, examining the circuit labels carefully: Assume the branches are: Left branch contains R1 and R2. Right branch contains R3 and R4. Node B is between R1 and R2. Node A is between R3 and R4. From the provided solution: This implies the branch containing node A consists of two equal resistances R, which contradicts the textual convention if R3 changed. Looking at the provided solution image logic, the branch with B has the changed resistance 1.1R. Let's follow the PDF's mathematical layout exactly. (Meaning branch A has equal resistors R and R) *(Note: The PDF solution computes VB = V / 2.1, implying the resistance below node B is R, and above is 1.1R)* The closest match among the options is 0.95 V. For a slightly unbalanced Wheatstone bridge due to a fractional change Delta R/R = x, the potential difference Delta V is approximately V cdot (x/4) for small x. Here x=0.1, so Delta V approx 40 x 0.1 / 4 = 1 V. Exact calculation gives 0.95 V. Class 12 Physics: Current Electricity

A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R1 = R2 = R3 = R4 ). When R3 resistance is heated to some temperature, its resistance v

Initially, R1 = R2 = R3 = R4 = R. The bridge is balanced, and VA = VB. After heating, R3 increases by 10%, meaning the new R3' = R + 0.1R = 1.1R. The other arms remain R. The battery provides a voltage of V = 40 V. Let the lower node (D) be at 0 V and the upper node (C) be at 40 V. The potential at node A (between R3' and R4) can be found via the voltage divider rule across the right branch (R3' and R4). Wait, examining the circuit labels carefully: Assume the branches are: Left branch contains

A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( R1 =…

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

Solution & Explanation

Core Logic

Step 1: Calculate Potentials at nodes A and B

Step 2: Potential Difference Calculation

Step 3: Final Answer

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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