Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 4.
Q46 (2024)
In the given circuit, the current in resistance $R_3$ is:
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- $1\text{ A}$
- $1.5\text{ A}$
- $2\text{ A}$
- $2.5\text{ A}$
### Related Formula
For parallel resistors:
$$R_{\text{parallel}} = \frac{R_a R_b}{R_a + R_b}$$
Total equivalent resistance in series:
$$R_{\text{eq}} = R_1 + R_{\text{parallel}} + R_4$$
Total current from source:
$$I = \frac{V}{R_{\text{eq}}}$$
### Core Logic
Analyzing the circuit network:
* $R_1 = 2\ \Omega$
* $R_2 = 4\ \Omega$ and $R_3 = 4\ \Omega$ are in parallel.
* $R_4 = 1\ \Omega$
Calculate the equivalent resistance of the parallel combination:
$$R_{\text{parallel}} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{4 \times 4}{4 + 4} = 2\ \Omega$$
### Step 1: Calculate Total Equivalent Resistance and Current
Total equivalent resistance is:
$$R_{\text{eq}} = R_1 + R_{\text{parallel}} + R_4 = 2 + 2 + 1 = 5\ \Omega$$
Total circuit current is:
$$I = \frac{V}{R_{\text{eq}}} = \frac{10\text{ V}}{5\ \Omega} = 2\text{ A}$$
### Step 2: Determine Current in R3
The total current of $2\text{ A}$ enters the parallel branch of $R_2$ and $R_3$. Since $R_2 = R_3 = 4\ \Omega$, the current divides equally between them:
$$I_{R_3} = I \times \frac{R_2}{R_2 + R_3} = 2 \times \frac{4}{8} = 1\text{ A}$$
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### Pattern Recognition
Equal parallel resistors split current exactly down the middle. Once total current is found to be $2\text{ A}$, the parallel branches share it as $1\text{ A}$ each without further math.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q58 (2024)
In the given circuit, the current flowing through the resistance $20\ \Omega$ is $0.3\text{ A}$, while the ammeter reads $0.9\text{ A}$. The value of $R_1$ is ________ $\Omega$.
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### Related Formula
For parallel branches, the potential difference $V$ across each branch is identical:
$V = I_i R_i$
According to Kirchhoff's Current Law, the total current $I_{\text{total}}$ is the sum of currents in all parallel branches:
$$I_{\text{total}} = i_1 + i_2 + i_3$$
### Core Logic
Analyzing the circuit diagram:
* Branch 1: Current $i_1$ through $20\ \Omega$ is $0.3\text{ A}$.
* Branch 2: Contains $15\ \Omega$ resistor with current $i_2$.
* Branch 3: Contains resistor $R_1$ with current $i_3$.
Since the branches are in parallel, they have the same potential difference $V_{AB}$:
$$V_{AB} = i_1 \times 20\ \Omega = 0.3\text{ A} \times 20\ \Omega = 6\text{ V}$$
### Step 1: Calculate Currents
Current through the second branch ($15\ \Omega$ resistor) is:
$$i_2 = \frac{V_{AB}}{15\ \Omega} = \frac{6\text{ V}}{15\ \Omega} = 0.4\text{ A}$$
Total current read by the ammeter is $0.9\text{ A}$. Thus:
$$i_1 + i_2 + i_3 = 0.9\text{ A}$$
$$0.3\text{ A} + 0.4\text{ A} + i_3 = 0.9\text{ A}$$
$$0.7\text{ A} + i_3 = 0.9\text{ A} \implies i_3 = 0.2\text{ A}$$
### Step 2: Calculate R1
Now use the voltage relation for the branch containing $R_1$:
$$i_3 \times R_1 = V_{AB}$$
$$(0.2\text{ A}) \times R_1 = 6\text{ V}$$
$$R_1 = \frac{6}{0.2} = 30\ \Omega$$
{{SOL_IMG1}}
### Pattern Recognition
In parallel networks, finding the branch voltage is always the primary step. Once $V = 6\text{ V}$ is established, the remaining branch currents are easily found using Ohm's Law and current conservation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q1559 (2024)
The charge accumulated on the capacitor connected in the following circuit is ______ $\mu\text{C}$. (Given $C = 150\ \mu\text{F}$).
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### Related Formula
$$Q = C \cdot \Delta V_C$$
### Core Logic
In steady state, no current flows through the capacitor branch. Analyze potential levels at node loops using standard Kirchhoff mesh values mapped across the core loop resistors:
$$V_A + \frac{10}{3}(1) - 6(1) = V_B$$
### Step 1: Compute node potential difference
$$V_A - V_B = 6 - \frac{10}{3} = \frac{8}{3}\text{ V}$$
### Step 2: Evaluate accumulated charge
$$Q = C(V_A - V_B) = 150\ \mu\text{F} \times \frac{8}{3}\text{ V} = 50 \times 8 = 400\ \mu\text{C}$$
### Pattern Recognition
Steady state capacitor loops act as open electrical cuts, transforming active mesh equations into simple layout node checks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q48 (2024)
A wire of length $10\text{ cm}$ and radius $\sqrt{7} \times 10^{-4}\text{ m}$ is connected across the right gap of a meter bridge. When a resistance of $4.5\ \Omega$ is connected on the left gap by using a resistance box, the balance length is found to be at $60\text{ cm}$ from the left end.
If the resistivity of the wire is $R \times 10^{-7}\ \Omega\text{m}$, then the value of $R$ is:
- $63$
- $70$
- $66$
- $35$
### Related Formula
From the balance condition of the meter bridge:
$$\frac{X_{\text{left}}}{l} = \frac{X_{\text{right}}}{100 - l}$$
Resistance formula:
$$X = \rho \frac{l_w}{A} = \frac{\rho l_w}{\pi r^2}$$
### Core Logic
First, evaluate the unknown resistance $X_{\text{right}}$ in the right gap:
$$\frac{4.5}{60} = \frac{X_{\text{right}}}{40} \implies X_{\text{right}} = \frac{4.5 \times 40}{60} = 3\ \Omega$$
### Step 1: Calculate Resistivity Value
Now map the resistance parameters ($l_w = 10\text{ cm} = 0.1\text{ m}$, $r = \sqrt{7} \times 10^{-4}\text{ m}$):
$$3 = \rho \frac{0.1}{\frac{22}{7} \times (\sqrt{7} \times 10^{-4})^2}$$
$$3 = \rho \frac{0.1}{\frac{22}{7} \times 7 \times 10^{-8}}$$
$$3 = \rho \frac{0.1}{22 \times 10^{-8}}$$
$$\rho = \frac{3 \times 22 \times 10^{-8}}{0.1} = 66 \times 10^{-7}\ \Omega\text{m}$$
### Step 2: Compare to find R
Given $\rho = R \times 10^{-7}$, comparing coefficients gives:
$R = 66$
### Pattern Recognition
Meter bridge balance simplifies directly to simple scalar component checks. The $\sqrt{7}$ term perfectly neutralizes the fractional $\frac{22}{7}$ constant in circular area profiles.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q49 (2024)
A wire of resistance $R$ and length $L$ is cut into $5$ equal parts. If these parts are joined parallely, then the resultant resistance will be:
- $\frac{1}{25}R$
- $\frac{1}{5}R$
- $25R$
- $5R$
### Core Logic
Resistance is directly proportional to length ($R \propto L$). Cutting the wire into 5 equal pieces reduces the resistance of each segment to:
$$R' = \frac{R}{5}$$
### Step 1: Compute parallel value
Connecting 5 identical resistors $R'$ in parallel gives a total equivalent resistance of:
$$R_{\text{eq}} = \frac{R'}{5} = \frac{R/5}{5} = \frac{R}{25}$$
### Pattern Recognition
Cutting an item into $N$ components and grouping them in parallel scales the overall baseline systemic value down cleanly by a factor of $N^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
