Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 3.
Q913 (2025)
Current passing through a wire as function of time is given as $I(t)=0.02t+0.01\mathrm{~A}$. The charge that will flow through the wire from $t=1\mathrm{~s}$ to $t=2\mathrm{~s}$ is:
- 0.06 C
- 0.02 C
- 0.07 C
- 0.04 C
### Related Formula
$$q = \int_{t_1}^{t_2} I(t) \, dt$$
### Core Logic
Given: $I(t) = (0.02t + 0.01)\text{ A}$
Limits: $t_1 = 1\text{ s}$, $t_2 = 2\text{ s}$
### Step 1: Perform Definite Integration
$$q = \int_{1}^{2} (0.02t + 0.01) \, dt$$
$$q = \left[ 0.02\frac{t^2}{2} + 0.01t \right]_{1}^{2} = \left[ 0.01t^2 + 0.01t \right]_{1}^{2}$$
$$q = \left[ 0.01(2)^2 + 0.01(2) \right] - \left[ 0.01(1)^2 + 0.01(1) \right]$$
$$q = [0.04 + 0.02] - [0.01 + 0.01] = 0.06 - 0.02 = 0.04\text{ C}$$
Hence, the total charge flowing through the wire is $0.04\text{ C}$.
### Pattern Recognition
Definite integration of a linear current function can also be verified geometrically by calculating the area of the trapezoid under the $I\text{--}t$ curve:
$$\text{Area} = \frac{I(1) + I(2)}{2} \times (2 - 1) = \frac{0.03 + 0.05}{2} \times 1 = 0.04\text{ C}$$
### Evaluation Rubric / Model Answer
Option D: 0.04 C
### Chapter Mix
Class 12 Physics: Current Electricity
Q23 (2025)
A wire of resistance $9\Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ______ ohm.
### Core Logic
The total resistance of the continuous uniform wire loop is $9\Omega$. When bent into an equilateral triangle, it is split into three equal length sections. The resistance of each individual side is:
$$R_{\text{side}} = \frac{9\Omega}{3} = 3\Omega$$
### Step 1: Calculating Equivalent Series and Parallel Resistance
As shown in the circuit diagrams {{SOL_IMG1}} and {{SOL_IMG2}}, measuring across any two vertices means one branch contains a single side resistor ($3\Omega$), while the other branch contains the remaining two sides connected in series :
$$R_{\text{series}} = 3\Omega + 3\Omega = 6\Omega$$
Now, calculate the parallel equivalent between these two branches:
$$R_{\text{eq}} = \frac{R_{\text{side}} \times R_{\text{series}}}{R_{\text{side}} + R_{\text{series}}} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\Omega$$
### Pattern Recognition
For a closed uniform loop with $N$ equal sides, the parallel resistance measured across adjacent corners always simplifies to $\frac{N-1}{N^2} \cdot R_{\text{total}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q24 (2025)
The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be ________ A.
{{IMG1}}
### Related Formula
For a balanced Wheatstone bridge network, if the potentials at opposite nodes are equal ($V_A = V_B$), no current flows through the central branch. The resistance arms satisfy the balance ratio:
$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$
Total current from the source is calculated using Ohm's law:
$$I = \frac{V_{\text{source}}}{R_{\text{equivalent}}}$$
### Core Logic
Given conditions :
$$V_A = V_B \implies \text{Balanced condition}$$
From the circuit network diagram [cite: 817, 818, 826, 828]:
* Left-top arm $= 10 \ \Omega$
* Left-bottom arm $= R$
* Right-top arm $= 20 \ \Omega$
* Right-bottom arm $= 40 \ \Omega$
Apply the balance condition to solve for unknown resistor $R$ :
$$\frac{10}{R} = \frac{20}{40} \implies \frac{10}{R} = \frac{1}{2} \implies R = 20 \ \Omega \quad \text{[cite: 837, 838]}$$
Now restructure the equivalent network :
Since the central $30 \ \Omega$ resistor branch carries zero current, it can be removed from the calculation [cite: 820, 834].
* Top \parallel branch: $10 \ \Omega + 20 \ \Omega = 30 \ \Omega$
* Bottom \parallel branch: $20 \ \Omega + 40 \ \Omega = 60 \ \Omega$
Calculate total equivalent resistance $R_{\text{eq}}$:
$$R_{\text{eq}} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20 \ \Omega$$
Calculate total current $I$ drawn from the $40\text{V}$ source:
$$I = \frac{40 \text{ V}}{20 \ \Omega} = 2 \text{ A}$$
### Step 1: Circuit Solution
The balanced bridge network layout with branch currents is shown below:
{{SOL_IMG1}}
### Pattern Recognition
When a question states that two nodes are at equal potential ($V_A = V_B$), immediately identify it as a balanced Wheatstone bridge. This allows you to remove the central branch and simplify the circuit into basic series-\parallel resistor combinations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q1354 (2024)
The reading in the ideal voltmeter (V) shown in the given circuit diagram is:
{{IMG1}}
- $5\mathrm{~V}$
- $10\mathrm{~V}$
- $0\mathrm{~V}$
- $3\mathrm{~V}$
### Related Formula
Total effective loop EMF and internal resistance:
$$I = \frac{E_{\text{net}}}{r_{\text{net}}}$$
Terminal potential difference across a discharging cell:
$V = E - Ir$
### Core Logic
The loop has a set of 8 cells all aiding the same current flow sequence direction.
$$E_{\text{eq}} = 8 \times 5 = 40\mathrm{~V}$$
$$r_{\text{eq}} = 8 \times 0.2 = 1.6\mathrm{~\Omega}$$
The circulating loop current is:
$$I = \frac{40\mathrm{~V}}{1.6\mathrm{~\Omega}} = 25\mathrm{~A}$$
### Step 1: Terminal Voltage Calculation
The ideal voltmeter measures the potential drop across the localized parallel branch configuration:
$$V = E - Ir = 5 - (25 \times 0.2) = 5 - 5 = 0\mathrm{~V}$$
### Pattern Recognition
A closed loop composed entirely of identical series active cells short circuits itself perfectly relative to the localized node drop points, reducing the net external voltage drop to zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q59 (2024)
The current in a conductor is expressed as $I = 3t^2 + 4t^3$, where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t = 1\mathrm{~s}$ to $t = 2\mathrm{~s}$ is _______ $\mathrm{C}$.
### Related Formula
Relationship between charge and time-varying current:
$$I = \frac{dq}{dt} \implies q = \int_{t_1}^{t_2} I \, dt$$
### Core Logic
Set up the definite integral using the given bounds $t=1\mathrm{~s}$ to $t=2\mathrm{~s}$:
$$q = \int_{1}^{2} (3t^2 + 4t^3) \, dt$$
Perform integration term-by-term:
$$q = \left[ \frac{3t^3}{3} + \frac{4t^4}{4} \right]_1^2 = \left[ t^3 + t^4 \right]_1^2$$
### Step 1: Evaluate Definite Bounds
Substitute upper and lower limits:
$$q = (2^3 + 2^4) - (1^3 + 1^4)$$
$$q = (8 + 16) - (1 + 1) = 24 - 2 = 22\mathrm{~C}$$
### Pattern Recognition
Simple polynomial integration. Always evaluate both boundary points explicitly to avoid dropped terms from the lower bound.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
