The reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the…

Year	2026	2025	2024	Total
Questions	18	13	19	50

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

C)

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 reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ____ A. A circuit diagram containing a 10V battery, multiple resistors, a 10 microfarad capacitor, and an ammeter.

Solution & Explanation

Core Logic

Step 1: Equivalent Resistance

Step 2: Ammeter Reading

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 reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ____ A.
A circuit diagram containing a 10V battery, multiple resistors, a 10 microfarad capacitor, and an ammeter.