Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 10.
Q48 (2026)
In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2 \Omega$ and $3 \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance l cm from the left. Now when an unknown resistance x $\Omega$ is connected in parallel to $3 \Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance x is ____ $\Omega$ .
### Related Formula
Meter bridge balance condition:
$$\frac{R_1}{R_2} = \frac{l}{100-l}$$
### Core Logic
In Case I:
$$\frac{2}{3} = \frac{\ell}{(100 - \ell)} \quad \dots (1)$$
### Step 1: Calculate initial null point
Solving Equation (1):
$$200 - 2\ell = 3\ell$$
$$5\ell = 200 \implies \ell = 40 \text{ cm}$$
### Step 2: Apply Case II conditions
In Case II, the right gap resistance becomes a parallel combination: $R' = \frac{3x}{3+x}$.
The new null point is shifted $10 \text{ cm}$ to the right, meaning the new balancing length is $\ell' = 40 + 10 = 50 \text{ cm}$.
$$\frac{2}{R'} = \frac{50}{100 - 50} = \frac{50}{50} = 1$$
Thus, $R' = 2 \, \Omega$.
### Step 3: Solve for x
$$\frac{3x}{3 + x} = 2$$
$3x = 6 + 2x$
$x = 6 \, \Omega$
### Pattern Recognition
If the null point falls exactly at $50 \text{ cm}$, the two gap resistances must be identical. Recognizing this immediately yields $R_{\text{parallel}} = 2$ without solving proportional fractions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q28 (2026)
In the potentiometer, when the cell in the secondary circuit is shunted with $4 \,\Omega$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell in shunted with $12 \,\Omega$ resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ____ $\Omega$.
- $3$
- $4$
- $12$
- $6$
### Related Formula
For a cell of emf $E$ and internal resistance $r$ shunted by a resistance $R$, the terminal voltage is:
$$V = \frac{E \cdot R}{r + R}$$
In a potentiometer, the terminal voltage is proportional to the balancing length $l$:
$V = K \cdot l$
### Core Logic
Equate the expressions for terminal voltage for both cases to their respective balancing lengths. Let $E$ be the emf and $r$ be the internal resistance of the cell.
{{SOL_IMG1}}
### Step 1: First Condition Setup
When shunted with $4 \,\Omega$:
$$\frac{E \cdot 4}{r + 4} = 120K \quad \text{--- (i)}$$
### Step 2: Second Condition Setup
When shunted with $12 \,\Omega$:
$$\frac{E \cdot 12}{r + 12} = 180K \quad \text{--- (ii)}$$
### Step 3: Solving for Internal Resistance
Divide equation (i) by equation (ii):
$$\frac{\frac{4E}{r + 4}}{\frac{12E}{r + 12}} = \frac{120K}{180K}$$
$$\frac{1}{3} \cdot \frac{r + 12}{r + 4} = \frac{2}{3}$$
$$r + 12 = 2(r + 4)$$
$r + 12 = 2r + 8$
$r = 4 \,\Omega$
### Pattern Recognition
Standard potentiometer problem. Ratio of balancing lengths maps directly to the ratio of terminal voltages. $\frac{l_1}{l_2} = \frac{R_1(r+R_2)}{R_2(r+R_1)}$. Plugging the values simplifies it instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q38 (2026)
For the two cells having same EMF E and internal resistance r, the current passing through the external resistor $6 \,\Omega$ is same when both the cells are connected either in parallel or in series. The value of internal resistance r is ____ $\Omega$.
- $3$
- $4$
- $9$
- $6$
### Related Formula
$$I_{\text{series}} = \frac{2E}{R + 2r}$$
$$I_{\text{parallel}} = \frac{E}{R + r/2}$$
### Core Logic
Since the external current is identical in both cases, equate the series current expression to the parallel current expression.
### Step 1: Setting up equation
$$\frac{2E}{6 + 2r} = \frac{E}{6 + \frac{r}{2}}$$
### Step 2: Solving for r
Divide both sides by $E$ and cross-multiply:
$$2 \left(6 + \frac{r}{2}\right) = 6 + 2r$$
$12 + r = 6 + 2r$
$12 - 6 = 2r - r$
$r = 6 \,\Omega$
### Pattern Recognition
A well-known standard shortcut: If $n$ identical cells yield the same current in series and parallel across an external resistor $R$, then $r = R$. Here $R = 6$, so immediately $r = 6$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q50 (2026)
The equivalent resistance between the points A and B in the following circuit is $\frac{x}{5}\Omega$ . The value of x is
{{IMG1}}
### Related Formula
$$\text{Kirchhoff's Voltage Law (KVL): } \sum V = 0$$
$$V = I \cdot R_{\text{eq}}$$
### Core Logic
This is an unbalanced Wheatstone bridge. We can solve it by connecting an imaginary 1V battery across AB, applying Kirchhoff's laws to find the total current $I$, and then finding $R_{\text{eq}} = V / I$.
{{SOL_IMG1}}
### Step 1: First KVL Loop
Let total current from the 1V source be $i$. It splits into $i_1$ through the upper $6\,\Omega$ resistor, and $(i - i_1)$ through the lower $3\,\Omega$ resistor.
Applying KVL to the upper closed loop of the bridge:
$$6i_1 + 3(2i_1 - i) = 3(i - i_1)$$
$$6i_1 + 6i_1 - 3i = 3i - 3i_1$$
$$15i_1 = 6i \Rightarrow i_1 = \frac{2}{5}i \quad \text{--- (1)}$$
### Step 2: Second KVL Loop (Outer Battery Loop)
Apply KVL through the lower path and the battery:
$$3(i - i_1) + 6(\text{current in next branch}) = 1$$
Wait, looking at the node splits in the solution: the current entering B from the bottom resistor is $i_1$, and from top is $(i - i_1)$ due to symmetry.
Equation from the lower path:
$$3(i - i_1) + 6i_1 = 1$$
Substitute $i_1 = \frac{2}{5}i$:
$3i + 3i_1 = 1$
$$3i + 3\left(\frac{2}{5}i\right) = 1$$
$$i \left(3 + \frac{6}{5}\right) = 1$$
$$i \left(\frac{21}{5}\right) = 1 \Rightarrow i = \frac{5}{21} \mathrm{~A}$$
### Step 3: Finding Equivalent Resistance
$$R_{\text{eq}} = \frac{V}{i} = \frac{1}{5 / 21} = \frac{21}{5} \,\Omega$$
Given $R_{\text{eq}} = \frac{x}{5} \,\Omega$.
Therefore, $x = 21$.
### Pattern Recognition
For a symmetrical unbalanced bridge (like $R_1=6, R_2=3$ and $R_3=3, R_4=6$), the currents divide symmetrically. Assuming a 1V source and tracking current $i$ and $i_1$ is the fastest fail-proof algebraic method.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q31 (2026)
A wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ( $R_{1} = R_{2} = R_{3} = R_{4}$ ). When $R_{3}$ resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference ( $V_{a} - V_{b}$ ) (after $R_{3}$ is heated) is ____ V.
{{IMG1}}
- $1.05$
- $0$
- $0.95$
- $2$
### Core Logic
{{SOL_IMG1}}
Initially, $R_{1} = R_{2} = R_{3} = R_{4} = R$. The bridge is balanced, and $V_A = V_B$.
After heating, $R_3$ increases by 10%, meaning the new $R_3' = R + 0.1R = 1.1R$.
The other arms remain $R$.
The battery provides a voltage of $V = 40\text{ V}$.
### Step 1: Calculate Potentials at nodes A and B
Let the lower node (D) be at $0\text{ V}$ and the upper node (C) be at $40\text{ V}$.
The potential at node A (between $R_3'$ and $R_4$) can be found via the voltage divider rule across the right branch ($R_3'$ and $R_4$).
Wait, examining the circuit labels carefully:
Assume the branches are: Left branch contains $R_1$ and $R_2$. Right branch contains $R_3$ and $R_4$.
Node B is between $R_1$ and $R_2$. Node A is between $R_3$ and $R_4$.
From the provided solution:
$$V_A = \frac{V}{2}$$
This implies the branch containing node A consists of two equal resistances $R$, which contradicts the textual convention if $R_3$ changed. Looking at the provided solution image logic, the branch with $B$ has the changed resistance $1.1R$.
Let's follow the PDF's mathematical layout exactly.
$$V_A = \frac{V}{2}$$ (Meaning branch A has equal resistors $R$ and $R$)
$$V_B = \frac{V}{R + 1.1R} \times R = \frac{V}{2.1}$$
*(Note: The PDF solution computes $V_B = V / 2.1$, implying the resistance below node B is $R$, and above is $1.1R$)*
### Step 2: Potential Difference Calculation
$$\therefore V_A - V_B = V \left[ \frac{1}{2} - \frac{1}{2.1} \right]$$
$$V_A - V_B = 40 \times \left[ \frac{2.1 - 2}{2 \times 2.1} \right]$$
$$V_A - V_B = \frac{0.1}{4.2} \times 40 = \frac{4}{4.2}$$
$$V_A - V_B = \frac{40}{42} \approx 0.952 \text{ V}$$
### Step 3: Final Answer
The closest match among the options is $0.95\text{ V}$.
### Pattern Recognition
For a slightly unbalanced Wheatstone bridge due to a fractional change $\Delta R/R = x$, the potential difference $\Delta V$ is approximately $V \cdot \frac{x}{4}$ for small $x$. Here $x=0.1$, so $\Delta V \approx 40 \times 0.1 / 4 = 1 \text{ V}$. Exact calculation gives $0.95\text{ V}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
