The equivalent resistance between the points A and B in the following circuit is (x/5)Omega . The value of x is

Year	2026	2025	2024	Total
Questions	18	13	19	50

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Kirchhoff's Voltage Law (KVL): Σ V = 0

V = I \cdot Req

Core Logic

This is an unbalanced Wheatstone bridge. We can solve it by connecting an imaginary 1V battery across AB, applying Kirchhoff's laws to find the total current $I$ $I$, and then finding $Req = V / I$ $R_{\text{eq}} = V / I$.

Circuit diagram with current loops drawn — A bridge network with resistors 6 Ohm, 3 Ohm in one path, a central 3 Ohm resistor, and 3 Ohm, 6 Ohm resistors in the other path between points A and B.

Step 1: First KVL Loop

Let total current from the 1V source be $i$ $i$. It splits into $i₁$ $i_1$ through the upper $6 Ω$ $6\,\Omega$ resistor, and $(i - i₁)$ $(i - i_1)$ through the lower $3 Ω$ $3\,\Omega$ resistor. Applying KVL to the upper closed loop of the bridge:

6i₁ + 3(2i₁ - i) = 3(i - i₁)

6i₁ + 6i₁ - 3i = 3i - 3i₁

15i₁ = 6i \Rightarrow i₁ = (2)/(5)i --- (1)

Step 2: Second KVL Loop (Outer Battery Loop)

Apply KVL through the lower path and the battery:

3(i - i₁) + 6(current in next branch) = 1

Wait, looking at the node splits in the solution: the current entering B from the bottom resistor is $i₁$ $i_1$, and from top is $(i - i₁)$ $(i - i_1)$ due to symmetry. Equation from the lower path:

3(i - i₁) + 6i₁ = 1

Substitute $i₁ = (2)/(5)i$ $i_1 = \frac{2}{5}i$: $3i + 3i₁ = 1$ $3i + 3i_1 = 1$

3i + 3((2)/(5)i) = 1

i (3 + (6)/(5)) = 1

i ((21)/(5)) = 1 \Rightarrow i = (5)/(21) ~A

Step 3: Finding Equivalent Resistance

Req = (V)/(i) = (1)/(5 / 21) = (21)/(5) Ω

Given $Req = (x)/(5) Ω$ $R_{\text{eq}} = \frac{x}{5} \,\Omega$. Therefore, $x = 21$ $x = 21$.

Pattern Recognition

For a symmetrical unbalanced bridge (like $R₁=6, R₂=3$ $R_1=6, R_2=3$ and $R₃=3, R₄=6$ $R_3=3, R_4=6$), the currents divide symmetrically. Assuming a 1V source and tracking current $i$ $i$ and $i₁$ $i_1$ is the fastest fail-proof algebraic method.

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

The equivalent resistance between the points A and B in the following circuit is (x)/(5)Ω$\frac{x}{5}\Omega$ . The value of x is A bridge network with resistors 6 Ohm, 3 Ohm in one path, a central 3 Ohm resistor, and 3 Ohm, 6 Ohm resistors in the other path between points A and B.

Solution & Explanation

Related Formula

Core Logic

Step 1: First KVL Loop

Step 2: Second KVL Loop (Outer Battery Loop)

Step 3: Finding Equivalent Resistance

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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The equivalent resistance between the points A and B in the following circuit is $(x)/(5)Ω$ $\frac{x}{5}\Omega$ . The value of x is
A bridge network with resistors 6 Ohm, 3 Ohm in one path, a central 3 Ohm resistor, and 3 Ohm, 6 Ohm resistors in the other path between points A and B.