Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 2.
Q15 (2025)
A wire of resistance $R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is
- $9 / 8$
- $8 / 9$
- 27/32
- 32/27
### Related Formula
$$\mathrm{R} = \frac{\rho\ell}{\mathrm{A}} \implies \mathrm{R} \propto \ell$$
### Core Logic
For the equilateral triangle, the total resistance $\mathrm{R}$ is divided into three equal line parts of value $\frac{\mathrm{R}}{3}$ each:
{{SOL_IMG1}}
{{SOL_IMG2}}
An edge configuration places one part in parallel with the other two series steps:
$$\left(\mathrm{R}_{\text{eq}}\right)_1 = \frac{(\frac{2\mathrm{R}}{3}) \cdot (\frac{\mathrm{R}}{3})}{\frac{2\mathrm{R}}{3} + \frac{\mathrm{R}}{3}} = \frac{2}{9}\mathrm{R}$$
For the square, the wire resistance $\mathrm{R}$ divides into four parts of $\frac{\mathrm{R}}{4}$ each:
{{SOL_IMG3}}
{{SOL_IMG4}}
An edge calculation pairs one section in parallel with the remaining three loops in series:
$$\left(\mathrm{R}_{\text{eq}}\right)_2 = \frac{(\frac{3\mathrm{R}}{4}) \cdot (\frac{\mathrm{R}}{4})}{\frac{3\mathrm{R}}{4} + \frac{\mathrm{R}}{4}} = \frac{3}{16}\mathrm{R}$$
Taking the ratio between these geometric layouts:
### Step 1: Final Ratio Calculation
$$\frac{\left(\mathrm{R}_{\text{eq}}\right)_1}{\left(\mathrm{R}_{\text{eq}}\right)_2} = \frac{\frac{2}{9}\mathrm{R}}{\frac{3}{16}\mathrm{R}} = \frac{32}{27}$$
This selects option (4).
### Pattern Recognition
For a regular polygon with $\mathrm{N}$ identical sides formed from a wire of resistance $\mathrm{R}$, the resistance across one edge is always given by $\frac{(\mathrm{N}-1)}{\mathrm{N}^2}\mathrm{R}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
Q5 (2025)
A wire of length $25\mathrm{~m}$ and cross-sectional area $5\mathrm{~mm}^2$ having resistivity of $2\times 10^{-6}\Omega\mathrm{~m}$ is bent into a complete circle. The resistance between diametrically opposite points will be:
- $12.5\Omega$
- $50\Omega$
- $100\Omega$
- $25\Omega$
### Related Formula
Resistance of a uniform wire:
$$R = \frac{\rho L}{A}$$
Equivalent resistance of two identical resistors in parallel:
$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{R_{\text{half}}}{2}$$
### Core Logic
First, find the total resistance $R_{\text{total}}$ of the straight wire:
- Length of wire, $L = 25\mathrm{~m}$
- Area of cross-section, $A = 5\mathrm{~mm}^2 = 5 \times 10^{-6}\mathrm{~m}^2$
- Resistivity, $\rho = 2 \times 10^{-6}\Omega\mathrm{~m}$
$$R_{\text{total}} = \frac{\rho L}{A} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10\Omega$$
### Step 1: Splitting into Semicircles
When the wire is bent into a complete circle, measuring the resistance between two diametrically opposite points splits the wire into two parallel halves of equal length.
Each semicircle has a resistance of:
$$R_{\text{semi}} = \frac{R_{\text{total}}}{2} = \frac{10}{2} = 5\Omega$$
These two halves are connected in parallel between the diametric terminals:
$$R_{\text{eq}} = \frac{R_{\text{semi}}}{2} = \frac{5}{2} = 2.5\Omega$$
{{SOL_IMG1}}
### Step 2: Conclusion & Discrepancy
The actual mathematically rigorous answer is $2.5\Omega$. Since $2.5\Omega$ is not present in the given options, the question is marked as a **Bonus** question by standard key evaluation guidelines. If forced to choose a theoretical option due to printing mistakes, some keys may relate it to $10\Omega / 4 = 2.5\Omega$, but scientifically it stands as a bonus.
### Pattern Recognition
Shortcut: A wire of total resistance $R$ bent into a circle has an effective resistance across its diameter equal to $R_{\text{eq}} = R/4$. Memorize this ratio! Here $R = 10\Omega$, so $R_{\text{eq}} = 10/4 = 2.5\Omega$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
Q25 (2025)
In the figure shown below, a resistance of $150.4\Omega$ is connected in series to an ammeter A of resistance $240\Omega$. A shunt resistance of $10\Omega$ is connected in parallel with the ammeter. The reading of the ammeter is ________ $\mathrm{mA}$.
{{IMG1}}
### Related Formula
Parallel combination of ammeter ($R_G = 240\Omega$) and shunt ($S = 10\Omega$):
$$R_{\text{parallel}} = \frac{R_G \cdot S}{R_G + S}$$
Total equivalent resistance:
$$R_{\text{eq}} = R_{\text{series}} + R_{\text{parallel}}$$
Current Divider Rule:
$$I_{\text{ammeter}} = I_{\text{total}} \cdot \left(\frac{S}{R_G + S}\right)$$
### Core Logic
Let's first calculate the equivalent resistance of the parallel branch (ammeter + shunt):
$$R_{\text{parallel}} = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6\Omega$$
Now, add the series resistance ($150.4\Omega$) to find the total equivalent resistance of the circuit:
$$R_{\text{eq}} = 150.4 + 9.6 = 160\Omega$$
### Step 1: Finding Total Current
Use Ohm's Law to calculate the total current $I_{\text{total}}$ drawn from the $20\mathrm{~V}$ battery:
$$I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{20}{160} = 0.125\mathrm{~A}$$
### Step 2: Calculating Ammeter Current
Using the Current Divider Rule, calculate the current $I_{\text{ammeter}}$ flowing through the $240\Omega$ ammeter branch:
$$I_{\text{ammeter}} = I_{\text{total}} \times \frac{10}{240 + 10} = 0.125 \times \frac{10}{250}$$
$$I_{\text{ammeter}} = 0.125 \times \frac{1}{25} = 0.005\mathrm{~A}$$
Convert this into milliamperes ($\mathrm{mA}$):
$$I_{\text{ammeter}} = 0.005 \times 1000 = 5\mathrm{~mA}$$
Therefore, the reading of the ammeter is $5\mathrm{~mA}$.
{{SOL_IMG1}}
{{SOL_IMG2}}
### Pattern Recognition
Notice how nicely the numbers are set up in JEE Mains questions! The parallel combination of $240\Omega$ and $10\Omega$ yields exactly $9.6\Omega$, which beautifully combines with $150.4\Omega$ to form a perfect integer sum of $160\Omega$. This tells you that your intermediate steps are absolutely correct!
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity: Measuring Devices
Q2 (2025)
There are 'n' number of identical electric bulbs, each is designed to draw a power p independently from the mains supply. They are now joined in series across the main supply. The total power drawn by the combination is:
- $np$
- $\frac{p}{n^{2}}$
- $\frac{p}{n}$
- $p$
### Related Formula
$$P = \frac{V^2}{R}$$
For series combination:
$$R_s = R_1 + R_2 + \dots + R_n$$
### Core Logic
Since the bulbs are identical, each has a resistance $R = \frac{V^2}{p}$.
When $n$ identical bulbs are connected in series, the total equivalent resistance becomes:
$R_s = nR$
### Step 1: Calculate Combined Power
The total power drawn across the same mains supply $V$ is:
$$P_s = \frac{V^2}{R_s} = \frac{V^2}{nR} = \frac{1}{n} \left(\frac{V^2}{R}\right) = \frac{p}{n}$$
### Pattern Recognition
Identical appliances connected in series scale down their combined power inversely with count ($P_{net} = P/n$), identical appliances in parallel scale up combined power linearly ($P_{net} = nP$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
Q6 (2025)
From the combination of resistors with resistance values $R_{1}=R_{2}=R_{3}=5\ \Omega$ and $R_{4}=10\ \Omega$, which of the following combination is the best circuit to get an equivalent resistance of $6\ \Omega$?
- {{IMG1}}
- {{IMG2}}
- {{IMG3}}
- {{IMG4}}
### Related Formula
Series combination:
$R_s = R_a + R_b$
Parallel combination:
$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_p = \frac{R_1 R_2}{R_1 + R_2}$$
### Core Logic
Let's check option (1):
Top branch has $R_1$ and $R_2$ in series: $R_{\text{top}} = 5 + 5 = 10\ \Omega$.
Bottom branch has $R_3$ and $R_4$ in series: $R_{\text{bottom}} = 5 + 10 = 15\ \Omega$.
### Step 1: Calculate Equivalent Parallel Resistance
These two branches are connected in parallel, so the total equivalent resistance is:
$$R_p = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \Omega$$
This matches the target value of $6\ \Omega$ perfectly. {{SOL_IMG1}}
### Pattern Recognition
Look for symmetric partitions. Standard combinations of values like $10\ \Omega$ and $15\ \Omega$ yield exactly $6\ \Omega$ in parallel.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
