Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics).
Q8 (2025)
The battery of a mobile phone is rated as $4.2\mathrm{~V}$, $5800\mathrm{~mAh}$. How much energy is stored in it when fully charged?
- $43.8\mathrm{~kJ}$
- $48.7\mathrm{~kJ}$
- $87.7\mathrm{~kJ}$
- $24.4\mathrm{~kJ}$
### Related Formula
$E = V \cdot Q$
$Q = I \cdot t$
### Core Logic
The capacity rating $5800\mathrm{~mAh}$ represents total electric charge $Q$ stored:
$$Q = 5800\mathrm{~mA} \times 1\mathrm{~hour} = (5800 \times 10^{-3}\mathrm{~A}) \times 3600\mathrm{~s} = 20880\mathrm{~C}$$
Given the voltage is $V = 4.2\mathrm{~V}$, the energy stored in Joules is:
$$E = V \cdot Q = 4.2 \times 20880 = 87696\mathrm{~J} = 87.696\mathrm{~kJ} \approx 87.7\mathrm{~kJ}$$
### Step 1: Final Conclusion
The electrical energy stored in the fully charged battery is $87.7\mathrm{~kJ}$.
### Pattern Recognition
Always remember: $\text{Energy (in Joules)} = \text{Voltage (V)} \times \text{Capacity (Ah)} \times 3600$. This directly transforms electrical rating units to standard SI energy units.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q173 (2025)
Two cells of emf 1V and 2V and internal resistance $2\Omega$ and $1\Omega$, respectively, are connected in series with an external resistance of $6\Omega$. The total current in the circuit is $I_{1}$ Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $I_{2}$. The value of $(\frac{I_{1}}{I_{2}})$ is $\frac{x}{3}$. The value of x is ________.
### Related Formula
- Series cell configuration:
$$\varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2, \quad r_{\text{eq}} = r_1 + r_2$$
- Parallel cell configuration (for unequal cells in parallel):
$$\varepsilon_{\text{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}, \quad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}$$
- Circuit current:
$$I = \frac{\varepsilon_{\text{eq}}}{r_{\text{eq}} + R}$$$
### Core Logic
Given parameters:
- Cell 1: $\varepsilon_1 = 1\mathrm{~V}$, $r_1 = 2\Omega$
- Cell 2: $\varepsilon_2 = 2\mathrm{~V}$, $r_2 = 1\Omega$
- External resistance $R = 6\Omega$
### Step 1: Calculate $I_1$ (Series Configuration){{SOL_IMG1}}
$$\varepsilon_{\text{eq}} = 1 + 2 = 3\mathrm{~V}$$
$$r_{\text{eq}} = 2 + 1 = 3\Omega$$
$$I_1 = \frac{3}{3 + 6} = \frac{3}{9} = \frac{1}{3}\mathrm{~A}$$$
### Step 2: Calculate $I_2$ (Parallel Configuration){{SOL_IMG2}}
$$\varepsilon_{\text{eq}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{0.5 + 2}{1.5} = \frac{2.5}{1.5} = \frac{5}{3}\mathrm{~V}$$
$$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\Omega$$
$$I_2 = \frac{\varepsilon_{\text{eq}}}{r_{\text{eq}} + R} = \frac{\frac{5}{3}}{\frac{2}{3} + 6} = \frac{\frac{5}{3}}{\frac{20}{3}} = \frac{5}{20} = \frac{1}{4}\mathrm{~A}$$$
### Step 3: Find ratio and evaluate $x$
$$\frac{I_1}{I_2} = \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{4}{3}$$
Comparing with $\frac{x}{3}$:
$x = 4$
### Pattern Recognition
For cells connected in parallel, the formula for equivalent EMF $\varepsilon_{\text{eq}}$ can be viewed as a weighted average. Connecting cells in series maximizes EMF but increases internal resistance, while parallel configuration limits EMF to an intermediate value while reducing equivalent internal resistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q3 (2025)
A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points A and B is R/n. The value of n is : {{IMG1}}
- 16
- 14
- 10
- 12
### Related Formula
For a wire of total resistance $R$ divided into $N$ equal segments, the resistance of each segment $r$ is:
$$r = \frac{R}{N}$$
For a balanced Wheatstone bridge with resistors of resistance $r$, the central arm can be neglected because no current flows through it.
### Core Logic
The triangular pyramid has $6$ segments of equal length. Since the total resistance of the wire is $R$:
$$r = \frac{R}{6}$$
Let the four vertices of the pyramid be $A$, $B$, $C$, and $D$. Terminals are at $A$ and $B$.
The segments are:
- $AB$ (direct path between terminals, resistance $r$)
- $AC$, $BC$, $AD$, $BD$ (forming a closed quadrilateral network between $A$ and $B$ with bridge arm $CD$)
- $CD$ (bridge arm connecting the midpoints, resistance $r$)
### Step 1: Simplify the Network
By symmetry, the potentials at $C$ and $D$ are equal when a voltage is applied across $A$ and $B$. Thus, the bridge is balanced, and no current flows through the segment $CD$.
We can remove segment $CD$ from the calculations:
- The path $A \to C \to B$ consists of two resistors in series: $r + r = 2r$.
- The path $A \to D \to B$ also consists of two resistors in series: $r + r = 2r$.
- The direct path $A \to B$ has a single resistor $r$.
These three parallel branches are connected between $A$ and $B$.
### Step 2: Calculate Equivalent Resistance
The equivalent resistance $R_{AB}$ is:
$$\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$$
$$R_{AB} = \frac{r}{2}$$
Substitute $r = \frac{R}{6}$:
$$R_{AB} = \frac{R/6}{2} = \frac{R}{12}$$
Comparing with $R_{AB} = \frac{R}{n}$, we find $n = 12$.
### Pattern Recognition
Sees: Resistor network formed by a 3D pyramid (6 identical edges, 4 nodes).
Shortcut: A 6-resistor regular tetrahedron has an equivalent resistance of $r/2$ across any two vertices. Since the total wire resistance is $R$ and it's cut into 6 pieces, $r = R/6 \implies R_{\text{eq}} = R/12$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q1 (2025)
The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have:
- $\text{low thermal conductivity and low electrical conductivity}$
- $\text{high thermal conductivity and high electrical conductivity}$
- $\text{low thermal conductivity and high electrical conductivity}$
- $\text{high thermal conductivity and low electrical conductivity}$
### Related Formula
$$V = S \cdot \Delta T$$
where,
$V$ = Thermoelectric voltage (Seebeck voltage)
$S$ = Seebeck coefficient
$\Delta T$ = Temperature difference
### Core Logic
To maximize the efficiency of a thermoelectric device harvesting heat energy, two conditions must be fulfilled:
1. **Low thermal conductivity**: This ensures that the temperature gradient ($\Delta T$) across the material is maintained and heat does not rapidly flow from the hot side to the cold side.
2. **High electrical conductivity**: This minimizes internal Joule heating losses ($I^2R$) when electrical current is drawn from the material.
Therefore, the material should possess low thermal conductivity and high electrical conductivity.
### Pattern Recognition
thermoelectric figure of merit is given by $Z = \frac{S^2 \sigma}{\kappa}$, where $\sigma$ is electrical conductivity and $\kappa$ is thermal conductivity. To maximize $Z$, we inherently need high $\sigma$ and low $\kappa$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Class 11 Physics: Thermal Properties of Matter
Q14 (2025)
Find the equivalent resistance between two ends of the following circuit.
{{IMG1}}
- $\mathrm{r}$
- $\frac{\mathrm{r}}{6}$
- $\frac{\mathrm{r}}{9}$
- $\frac{\mathrm{r}}{3}$
### Core Logic
Label the circuit nodes carefully to see how the elements are connected across terminal zones:
{{SOL_IMG1}}
{{SOL_IMG2}}
Tracing parallel loops reveals all inner subsets share identical path boundaries.
$$\mathrm{R}_{\text{eq}} = \frac{\mathrm{r}/3}{3} = \frac{\mathrm{r}}{9}$$
### Step 1: Final Expression
The overall simplified system network value equals $\frac{\mathrm{r}}{9}$, matching option (3).
### Pattern Recognition
Short-circuit tracking lines allow collapsing complex meshes into basic parallel branches. If all nodes connect symmetrically, apply simple division: $\mathrm{R}/\mathrm{N}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
