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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Solution for Wheatstone Bridge — A standard wheatstone bridge circuit powered by a 40V source.

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

Step 1: Calculate Potentials at nodes A and B

Let the lower node (D) be at $0 V$ $0\text{ V}$ and the upper node (C) be at $40 V$ $40\text{ V}$. The potential at node A (between $R₃'$ $R_3'$ and $R₄$ $R_4$) can be found via the voltage divider rule across the right branch ( $R₃'$ $R_3'$ and $R₄$ $R_4$).

Wait, examining the circuit labels carefully: Assume the branches are: Left branch contains $R₁$ $R_1$ and $R₂$ $R_2$. Right branch contains $R₃$ $R_3$ and $R₄$ $R_4$. Node B is between $R₁$ $R_1$ and $R₂$ $R_2$. Node A is between $R₃$ $R_3$ and $R₄$ $R_4$.

From the provided solution:

VA = (V)/(2)

This implies the branch containing node A consists of two equal resistances $R$ $R$, which contradicts the textual convention if $R₃$ $R_3$ changed. Looking at the provided solution image logic, the branch with $B$ $B$ has the changed resistance $1.1R$ $1.1R$. Let's follow the PDF's mathematical layout exactly.

$VA = (V)/(2)$ $$V_A = \frac{V}{2}$$ (Meaning branch A has equal resistors $R$ $R$ and $R$ $R$)

VB = (V)/(R + 1.1R) \times R = (V)/(2.1)

(Note: The PDF solution computes $VB = V / 2.1$ $V_B = V / 2.1$, implying the resistance below node B is $R$ $R$, and above is $1.1R$ $1.1R$)

Step 2: Potential Difference Calculation

VA - VB = V [ (1)/(2) - (1)/(2.1) ]

VA - VB = 40 \times [ (2.1 - 2)/(2 \times 2.1) ]

VA - VB = (0.1)/(4.2) \times 40 = (4)/(4.2)

VA - VB = (40)/(42) \approx 0.952 V

Step 3: Final Answer

The closest match among the options is $0.95 V$ $0.95\text{ V}$.

Pattern Recognition

For a slightly unbalanced Wheatstone bridge due to a fractional change $Δ R/R = x$ $\Delta R/R = x$, the potential difference $Δ V$ $\Delta V$ is approximately $V \cdot (x)/(4)$ $V \cdot \frac{x}{4}$ for small $x$ $x$. Here $x=0.1$ $x=0.1$, so $Δ V \approx 40 \times 0.1 / 4 = 1 V$ . Exact calculation gives $0.95 V$ $0.95\text{ V}$.

Chapter Mix

Class 12 Physics: Current Electricity

More Current Electricity Previous-Year Questions — Page 2

Q36 jee_main_2026_22_january_evening Power Transmission and Efficiency

An electric power line having total resistance of

2 Ω

, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is ____.

A. $96.9$ $96.9$
B. $86.5$ $86.5$
C. $100$ $100$
D. $92.5$ $92.5$

Solution

Related Formula

Pout = V \cdot I

Ploss = I² R

η = ( PoutPₙₑₜ) \times 100%

Core Logic

Calculating total current $I$ $I$:

1000 = 250 \times I I = 4 ~A

Calculating power loss along the line $Ploss$ $P_{\text{loss}}$:

Ploss = I² R = (4)² \times 2 = 32 ~W

Total input power supplied to line $Pₙₑₜ$ $P_{\text{net}}$:

Pₙₑₜ = Pout + Ploss = 1000 + 32 = 1032 ~W

Calculating transmission efficiency $η$ $\eta$:

η = ((1000)/(1032)) \times 100% \approx 96.9%

Step 1: Final Conclusion

The percentage efficiency of the transmission line is $96.9%$ $96.9\%$.

Pattern Recognition

Efficiency formula: $η = PoutPout + I² R \times 100%$ . Current $I = 1000/250 = 4~A$ $I = 1000/250 = 4\mathrm{~A}$. Loss $= 16 \times 2 = 32~W$ $= 16 \times 2 = 32\mathrm{~W}$. $η = 1000/1032 = 96.9%$ $\eta = 1000/1032 = 96.9\%$.

Chapter Mix

Class 12 Physics: Current Electricity

Q49 jee_main_2026_22_january_evening Drift Velocity and Electron Mobility

A cylindrical conductor of length 2m and area of cross-section

0.2 ~mm²

carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is

α \times 10⁻³ ~m²/V\dots

. The value of

α

is : (electron concentration =

5 \times 10²⁸/m³

and electron charge =

1.6 \times 10⁻¹⁹

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

$I = n e A vd$ $I = n e A v_d$

vd = μ E = μ (V)/(l)

μ = (I l)/(n e A V)

Core Logic

Combining current density and mobility equations:

I = n e A (μ (V)/(l)) μ = (I \cdot l)/(n \cdot e \cdot A \cdot V)

Substituting given values $I = 1.6 ~A, l = 2 ~m, n = 5 \times 10²⁸ /m³, e = 1.6 \times 10⁻¹⁹ ~C, A = 0.2 \times 10⁻⁶ ~m², V = 2 ~V$ :

μ = 1.6 \times 2(5 \times 10²⁸) \times (1.6 \times 10⁻¹⁹) \times (0.2 \times 10⁻⁶) \times 2

μ = (3.2)/(1.6 \times 10³ \times 2) = (3.2)/(3200) = 1.0 \times 10⁻³ ~m²/V\dots

Comparing with $α \times 10⁻³ α = 1$ .

Step 1: Final Conclusion

The value of $α$ $\alpha$ is 1.

Pattern Recognition

Mobility formula: $μ = I l / (n e A V)$ $\mu = I l / (n e A V)$. Direct parameter plug-in yields $α = 1$ $\alpha = 1$.

Chapter Mix

Class 12 Physics: Current Electricity

Q34 jee_main_2026_23_january_morning Resistance and Ohm's Law

A wire of uniform resistance

λΩ/m

is bent into a circle of radius r and another piece of wire with length 2r is connected between points A and B (AOB) as shown in figure. The equivalent resistance between points A and B is ____

Ω

Resistance and Ohm's Law diagram for Q34 - JEE Main 2026 Morning — A ring with nodes A and B connected across a diameter forming parallel branches.

A. $(3πλ r)/(8)$ $\frac{3\pi\lambda r}{8}$
B. $(π + 1)2rλ$ $(\pi + 1)2r\lambda$
C. $(6πλ r)/(3π + 16)$ $\frac{6\pi\lambda r}{3\pi + 16}$
D. $2π λ r$ $2\pi \lambda r$

Solution

Related Formula

$R = ρ L$ $R = \rho L$ For parallel resistors:

1Req = 1R₁ + 1R₂ + 1R₃

Core Logic

The system represents three resistors connected in parallel between nodes A and B: the upper arc, the lower arc, and the straight diameter wire. Using $R = λ L$ $R = \lambda L$ where $λ$ $\lambda$ is resistance per unit length.

Step 1: Assign Resistances

Length of the upper arc = $(π r)/(2)$ $\frac{\pi r}{2}$ Resistance $R₁ = λ \cdot (π r)/(2)$ $R_{1} = \lambda \cdot \frac{\pi r}{2}$

Length of the straight wire AB = $2r$ $2r$ Resistance $R₂ = λ \cdot 2r$ $R_{2} = \lambda \cdot 2r$

Length of the remaining larger arc = $2π r - (π r)/(2) = (3π r)/(2)$ Resistance $R₃ = λ \cdot (3π r)/(2)$

Step 2: Equivalent Resistance Calculation

1RAB = 1R₁ + 1R₂ + 1R₃

1RAB = (2)/(λ π r) + (1)/(2λ r) + (2)/(3λ π r)

1RAB = (1)/(λ r)[(2)/(π) + (1)/(2) + (2)/(3π)]

1RAB = (1)/(λ r)((12 + 3π + 4)/(6π)) = (1)/(λ r)((16 + 3π)/(6π))

Step 3: Final Inversion

RAB = λ r ((6π)/(16 + 3π))

Pattern Recognition

Sees: "Uniform resistance wire bent into shape" → immediately break the shape down into parallel/series segments defined purely by their arc/line lengths multiplied by the linear density $λ$ $\lambda$.

Chapter Mix

Class 12 Physics: Current Electricity

Q35 jee_main_2026_23_january_evening Measuring Instruments

To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is ____

A. $1.45$ $1.45$
B. $1.65$ $1.65$
C. $1.75$ $1.75$
D. $1.55$ $1.55$

Solution

Related Formula

ε = λ l

(Δ y)/(y) = (Δ l₁)/(l₁) + (Δ l₂)/(l₂)

Core Logic

By potentiometer principle: $ε₁ = λ l₁$ $\epsilon_1 = \lambda l_1$ $ε₂ = λ l₂$ $\epsilon_2 = \lambda l_2$

Ratio $y = (ε₁)/(ε₂) = (l₁)/(l₂)$ Maximum percentage error is the sum of fractional errors.

Step 1: Calculate Percentage Error

(Δ y)/(y) = (1)/(200) + (1)/(150)

(Δ y)/(y) = (3 + 4)/(600) = (7)/(600)

Percentage error = $(7)/(600) \times 100% = (7)/(6)% \approx 1.16%$

Pattern Recognition

Note: The calculated exact percentage error is 1.16%, which does not match any of the provided options exactly. The closest official option provided was tracked as index (2) marking 1.65 by some keys, though technically a bonus question.

Chapter Mix

Class 12 Physics: Current Electricity Class 11 Physics: Units and Measurements

Q30 jee_main_2026_24_january_morning Measuring Instruments

Two resistors of

100 Ω

each are connected in series with a 9V battery. A voltmeter of

400Ω

resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ____ V.

A. $3$ $3$
B. $4.5$ $4.5$
C. $4$ $4$
D. $2$ $2$

Solution

Related Formula

Req(parallel) = (R₁ R₂)/(R₁ + R₂)

$V = I \cdot R$ $V = I \cdot R$

Core Logic

Circuit diagram with voltmeter in parallel

The voltmeter is connected in parallel with one of the $100 Ω$ $100\, \Omega$ resistors. The equivalent resistance of this parallel combination is:

Rparallel = (400 \times 100)/(400 + 100) = (40000)/(500) = 80 Ω

The total equivalent resistance of the circuit is:

Req = 100 + 80 = 180 Ω

Step 1: Circuit Current and Voltmeter Reading

Current in the circuit:

I = EReq = (9)/(180) = (1)/(20) A

The reading of the voltmeter is the voltage across the parallel combination:

V = I \times 80 = (1)/(20) \times 80 = 4 V

Pattern Recognition

When a real voltmeter is used, it draws current. Model it as a resistor in parallel with the test component to find the exact altered potential drop.

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 2

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Assign Resistances

Step 2: Equivalent Resistance Calculation

Step 3: Final Inversion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Percentage Error

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Circuit Current and Voltmeter Reading

Pattern Recognition

Chapter Mix

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