Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 7.
Q2003 (2024)
Equivalent resistance of the following network is ________ $\Omega$
{{IMG1}}
### Related Formula
$$R_{\text{parallel}} = \frac{1}{\sum \frac{1}{R_i}}$$
### Core Logic
{{SOL_IMG1}}
By carefully identifying the nodes, we can see that a $6\,\Omega$ resistor in the middle branch is short-circuited by a direct zero-resistance wire path across it.
{{SOL_IMG2}}
Removing the short-circuited $6\,\Omega$ resistor simplifies the circuit into three identical branches connected directly between the terminals $A$ and $B$.
{{SOL_IMG3}}
### Step 2: Equivalent Calculation
The simplified circuit consists of three identical $3\,\Omega$ resistors in parallel.
$$R_{\text{eq}} = 3 \times \frac{1}{3} = 1\,\Omega$$
### Pattern Recognition
Always trace nodes directly connected by straight wires (zero resistance). Any resistor with both ends connecting to the exact same electrical node is shorted out and can be erased.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
Q48 (2024)
Two conductors have the same resistances at $0^{\circ}\mathrm{C}$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients for their series and parallel combinations are :
- $\alpha_1 + \alpha_2, \frac{\alpha_1 + \alpha_2}{2}$
- $\frac{\alpha_1 + \alpha_2}{2}, \frac{\alpha_1 + \alpha_2}{2}$
- $\alpha_1 + \alpha_2, \frac{\alpha_1\alpha_2}{\alpha_1 + \alpha_2}$
- $\frac{\alpha_1 + \alpha_2}{2}, \alpha_1 + \alpha_2$
### Related Formula
$$R_T = R_0(1 + \alpha \Delta T)$$
### Step 1: Series Combination
Let base resistance at $0^\circ \text{C}$ be $R$.
For series:
$$R_{\text{eq}} = R_1 + R_2$$
$$(2R)[1 + \alpha_{\text{eq,s}} \Delta T] = R(1 + \alpha_1 \Delta T) + R(1 + \alpha_2 \Delta T)$$
$$2 + 2\alpha_{\text{eq,s}} \Delta T = 2 + (\alpha_1 + \alpha_2)\Delta T$$
$$\alpha_{\text{eq,s}} = \frac{\alpha_1 + \alpha_2}{2}$$
### Step 2: Parallel Combination
For parallel at $0^\circ \text{C}$, $R_{\text{eq,0}} = R/2$.
$$R_{\text{eq,p}} = \frac{R_1 R_2}{R_1 + R_2}$$
$$\frac{R}{2} [1 + \alpha_{\text{eq,p}} \Delta T] = \frac{R^2 (1 + \alpha_1 \Delta T)(1 + \alpha_2 \Delta T)}{R(2 + (\alpha_1 + \alpha_2)\Delta T)}$$
$$\frac{1}{2} (1 + \alpha_{\text{eq,p}} \Delta T) = \frac{1 + (\alpha_1 + \alpha_2)\Delta T}{2\left(1 + \frac{\alpha_1 + \alpha_2}{2}\Delta T\right)}$$
Using binomial expansion for small $\Delta T$:
$$1 + \alpha_{\text{eq,p}} \Delta T \approx [1 + (\alpha_1 + \alpha_2)\Delta T] \left[ 1 - \frac{\alpha_1 + \alpha_2}{2}\Delta T \right]$$
$$1 + \alpha_{\text{eq,p}} \Delta T \approx 1 + (\alpha_1 + \alpha_2)\Delta T - \frac{\alpha_1 + \alpha_2}{2}\Delta T$$
$$\alpha_{\text{eq,p}} \Delta T = \frac{\alpha_1 + \alpha_2}{2}\Delta T$$
$$\alpha_{\text{eq,p}} = \frac{\alpha_1 + \alpha_2}{2}$$
### Pattern Recognition
For two identical base resistances, the effective temperature coefficient is simply the arithmetic mean of their individual coefficients, regardless of whether they are in series or parallel.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
Q2092 (2026)
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9V with internal resistance of $1 \Omega$ is connected across these points is ____ J.
{{IMG1}}
### Related Formula
$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \text{ (parallel)}$$
$H = I^2 R t$
$$I = \frac{V}{R_{\text{eq}} + r}$$
### Core Logic
The circuit is a balanced Wheatstone bridge between A and B, because the ratio of adjacent arms is $\frac{1}{2} = \frac{2}{4}$. Thus, the middle $1\,\Omega$ resistor (if there is one connecting the middle nodes) is ineffective.
{{SOL_IMG1}}
The equivalent resistance across AB ($R_{AB}$) involves the upper branch ($1+2 = 3\,\Omega$) and the lower branch ($2+4 = 6\,\Omega$) in parallel.
{{SOL_IMG2}}
$$R_{AB} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$$
### Step 1: Calculate Current and Heat
Total resistance in the circuit = $R_{AB} + r_{\text{internal}} = 2 + 1 = 3\,\Omega$.
Total current from the battery $i = \frac{9\text{V}}{3\,\Omega} = 3\text{A}$.
The heat generated specifically between points A and B in 1 minute (60 seconds) is:
$$H_{AB} = i^2 R_{AB} t = (3)^2 \times 2 \times 60$$
$$H_{AB} = 9 \times 120 = 1080\text{ J}$$
### Pattern Recognition
Check for balanced Wheatstone bridge first. Then use $H = i^2 R t$ strictly with the equivalent resistance of just the section $AB$ to find heat specifically generated across $AB$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
Q28 (2026)
The total length of potentiometer wire AB is $50 \text{ cm}$ in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is ________ cm.
{{IMG1}}
- $15$
- $30$
- $25$
- $20$
### Related Formula
For a balanced Wheatstone bridge configuration:
$$\frac{R_1}{R_2} = \frac{l_1}{l_2}$$
### Core Logic
When the galvanometer shows zero reading, it implies that the bridge is balanced. The resistance of the wire segments is proportional to their lengths.
Let the length of segment AP be $l_{AP}$ and PB be $l_{PB}$.
$$\frac{6\,\Omega}{R_{AP}} = \frac{4\,\Omega}{R_{PB}}$$
Since $R \propto l$:
$$\frac{R_{AP}}{R_{PB}} = \frac{l_{AP}}{l_{PB}} = \frac{6}{4} = \frac{3}{2}$$
### Step 1: Applying Total Length Constraint
We are given the total length:
$$l_{AP} + l_{PB} = 50 \text{ cm}$$
Using the ratio $l_{AP} = \frac{3}{2} l_{PB}$:
$$\frac{3}{2}l_{PB} + l_{PB} = 50$$
$$\frac{5}{2}l_{PB} = 50 \implies l_{PB} = 20 \text{ cm}$$
### Step 2: Final Conclusion
Solving for $l_{AP}$:
$$l_{AP} = 50 - 20 = 30 \text{ cm}$$
Alternatively, $l_{AP} = \frac{3}{3+2} \times 50 = \frac{3}{5} \times 50 = 30 \text{ cm}$.
### Pattern Recognition
Null deflection over a uniform wire directly translates resistance ratios into length ratios. Standard Wheatstone bridge application wrapped in potentiometer geometry.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
Q35 (2026)
A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when :
- $R=2r$
- $R=\frac{r}{2}$
- $R=\sqrt{2}r$
- $R=r$
### Related Formula
$$P = I^2 R = \left(\frac{E}{R+r}\right)^2 R$$
### Core Logic
To maximize power across the load resistance $R$, we apply the Maximum Power Transfer Theorem. The power transferred to the load is maximum when the load resistance equals the internal resistance of the battery.
Setting $\frac{dP}{dR} = 0$ yields:
$R = r$
### Step 1: Final Conclusion
Power is maximum when the external resistance is strictly equal to the internal resistance ($R=r$).
### Pattern Recognition
Direct theoretical application of the Maximum Power Transfer Theorem.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
