Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 6.
Q33 (2024)
A potential divider circuit is shown in figure. The output voltage $V_0$ is
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- $4 \mathrm{~V}$
- $2 \mathrm{~mV}$
- $0.5 \mathrm{~V}$
- $12 \mathrm{~mV}$
### Related Formula
$V = IR$
$$R_{\text{eq}} = R_1 + R_2 + \dots + R_n \quad (\text{for series})$$
### Core Logic
Observe the circuit diagram. The total equivalent resistance $R_{\text{eq}}$ of the series network must be calculated by summing all the resistance values shown in the main branch.
### Step 1: Calculate Total Resistance and Current
From the given network, assuming the total series resistance is $4000 \,\Omega$ (comprising the $3.3\mathrm{k}\Omega$ resistor and seven $100\,\Omega$ resistors).
$$R_{\text{eq}} = 4000 \,\Omega$$
The total voltage applied across the network is $4 \mathrm{~V}$.
$$i = \frac{V}{R_{\text{eq}}} = \frac{4}{4000} = \frac{1}{1000} \mathrm{~A}$$
### Step 2: Calculate Output Voltage
The output voltage $V_0$ is tapped across five $100 \,\Omega$ resistors.
$$R_{\text{out}} = 5 \times 100 \,\Omega = 500 \,\Omega$$
Thus, the output voltage is:
$$V_0 = i \cdot R_{\text{out}} = \left(\frac{1}{1000}\right) \times 500 = 0.5 \mathrm{~V}$$
### Pattern Recognition
A potential divider simply scales the input voltage by the fraction of the resistance tapped over the total resistance: $V_0 = V_{in} \times (R_{\text{tap}} / R_{\text{total}})$. Standard DC circuit division.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q43 (2024)
An electric toaster has resistance of $60\Omega$ at room temperature $(27^{\circ}\mathrm{C})$. The toaster is connected to a $220\mathrm{V}$ supply. If the current flowing through it reaches $2.75\mathrm{A}$, the temperature attained by toaster is around: (if $\alpha = 2\times 10^{-4} / ^{\circ}\mathrm{C}$)
- $694^{\circ} \mathrm{C}$
- $1235^{\circ}\mathrm{C}$
- $1694^{\circ}\mathrm{C}$
- $1667^{\circ}\mathrm{C}$
### Related Formula
$$R = \frac{V}{I}$$
$$R = R_0 (1 + \alpha \Delta T)$$
### Core Logic
First, evaluate the final resistance $R_T$ at the operating condition using Ohm's law. Second, plug the final resistance into the linear temperature dependence equation for resistance to solve for final temperature $T$.
### Step 1: Calculate Final Resistance
Given $V = 220 \mathrm{~V}$ and $I = 2.75 \mathrm{~A}$:
$$R_T = \frac{220}{2.75} = 80 \,\Omega$$
### Step 2: Apply Temperature Equation
We know $R_{27} = 60 \,\Omega$.
$$R_T = R_{27} [1 + \alpha (T - 27)]$$
$$80 = 60 [1 + 2 \times 10^{-4} (T - 27)]$$
$$\frac{80}{60} = 1 + 2 \times 10^{-4} (T - 27)$$
$$\frac{4}{3} - 1 = 2 \times 10^{-4} (T - 27)$$
$$\frac{1}{3} = 2 \times 10^{-4} (T - 27)$$
### Step 3: Solve for T
$$T - 27 = \frac{1}{6 \times 10^{-4}}$$
$$T - 27 = \frac{10000}{6} = 1666.67$$
$$T = 1666.67 + 27 \approx 1693.67 \Rightarrow 1694^\circ \mathrm{C}$$
### Pattern Recognition
Always separate the final temperature $T$ from $\Delta T$. The most common error is forgetting to add back the initial reference temperature ($27^\circ\mathrm{C}$) at the end.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q39 (2024)
The resistance per centimeter of a meter bridge wire is $r$, with $X \, \Omega$ resistance in left gap. Balancing length from left end is at $40 \text{ cm}$ with $25 \, \Omega$ resistance in right gap. Now the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for same settings will be at
- $20 \text{ cm}$
- $10 \text{ cm}$
- $80 \text{ cm}$
- $40 \text{ cm}$
### Related Formula
$$\frac{R_{\text{left}}}{R_{\text{wire-left}}} = \frac{R_{\text{right}}}{R_{\text{wire-right}}}$$
### Core Logic
For a meter bridge, the balancing condition is independent of the absolute resistance of the bridge wire as long as it is uniform. The ratio of the resistances in the gaps balances with the ratio of lengths.
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### Step 1: First Condition
$$\frac{X}{r \ell_1} = \frac{25}{r (100 - \ell_1)}$$
Given $\ell_1 = 40 \text{ cm}$:
$$\frac{X}{r \times 40} = \frac{25}{r \times 60} \implies \frac{X}{40} = \frac{25}{60}$$
### Step 2: Second Condition
When replaced by a wire of $2r$ per cm, the new lengths $\ell_2$ will satisfy:
$$\frac{X}{2r \ell_2} = \frac{25}{2r (100 - \ell_2)}$$
Notice that the $2r$ terms cancel out entirely from both sides, leaving:
$$\frac{X}{\ell_2} = \frac{25}{100 - \ell_2}$$
### Step 3: Conclusion
Since the ratio $X/25$ remains identical, the balancing length ratio $\ell / (100-\ell)$ also remains identical.
Therefore, $\ell_2 = \ell_1 = 40 \text{ cm}$.
### Pattern Recognition
Meter bridge balance point strictly depends on length ratio, NOT the specific resistivity or thickness of the wire (provided it is uniform). If external resistors don't change, the balance point never changes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q43 (2024)
By what percentage will the illumination of the lamp decrease if the current drops by $20\%$?
- $46\%$
- $26\%$
- $36\%$
- $56\%$
### Related Formula
Power (Illumination) is proportional to the square of the current for a constant resistance:
$P = I^2 R$
### Core Logic
Initial power:
$P_1 = I_1^2 R$
If current drops by 20%, the new current is:
$$I_2 = I_1 - 0.2 I_1 = 0.8 I_1$$
New power:
$$P_2 = (0.8 I_1)^2 R = 0.64 I_1^2 R = 0.64 P_1$$
### Step 1: Calculate Percentage Change
$$\Delta P \% = \frac{P_2 - P_1}{P_1} \times 100\%$$
$$\Delta P \% = \frac{0.64 P_1 - P_1}{P_1} \times 100\%$$
$$\Delta P \% = (0.64 - 1) \times 100\% = -36\%$$
### Step 2: Final Statement
The negative sign indicates a decrease. The illumination drops by $36\%$.
### Pattern Recognition
For squared relations $y = x^2$, if $x$ changes by a factor $k$ ($0.8$), $y$ changes by a factor $k^2$ ($0.64$). $1 - 0.64 = 36\%$. This bypasses algebraic limits usually done for small changes (like $2\Delta x / x$) since 20% is too large for approximation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q51 (2024)
In the following circuit, the battery has an emf of $2 \text{ V}$ and an internal resistance of $\frac{2}{3} \, \Omega$. The power consumption in the entire circuit is ______ W.
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### Related Formula
$$P = \frac{V^2}{R_{eq}}$$
### Core Logic
To find total power, collapse the entire external circuit and battery internal resistance into a single equivalent resistance $R_{eq}$ across the battery's ideal terminals.
### Step 1: Equivalent Resistance Calculation
Analyzing the diagram:
The circuit simplifies to an equivalent resistance $R_{eq}$ combining the parallel/series elements along with the internal resistance $r = 2/3 \, \Omega$.
The final simplified equivalent resistance of the entire system calculates to:
$$R_{eq} = \frac{4}{3} \, \Omega$$
### Step 2: Calculate Power
$$P = \frac{V^2}{R_{eq}}$$
$$P = \frac{2^2}{4/3}$$
$$P = \frac{4}{4/3} = 3 \text{ W}$$
### Pattern Recognition
Whenever "entire circuit" power is asked, include the battery's internal resistance inside $R_{eq}$ so you can use $P = E^2 / R_{total}$ directly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Current Electricity
