Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 5.
Q1647 (2024)
A $16 \, \Omega$ wire is bend to form a square loop. A 9V battery with internal resistance $1 \, \Omega$ is connected across one of its sides. If a $4 \, \mu \mathrm{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\frac{x}{2} \, \mu \mathrm{J}$. where $x = $ _________.
### Related Formula
Energy stored ($U$) by a capacitor of capacitance $C$ under steady state voltage $V_C$:
$$U = \frac{1}{2} C V_C^2$$
### Core Logic
A wire of resistance $16 \, \Omega$ is bent into a square, so each of the 4 sides has a resistance of:
$$R_{\text{side}} = \frac{16}{4} = 4 \, \Omega$$
The battery is connected across one side. Let this side have resistance $4 \, \Omega$. The remaining three sides are connected in series, creating a parallel branch with a combined resistance of:
$$R_{\text{series}} = 4 + 4 + 4 = 12 \, \Omega$$
{{SOL_IMG1}}
### Step 1: Calculate Equivalent Resistance and Circuit Current
The equivalent external resistance of the parallel loop branches is:
$$R_p = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 \, \Omega$$
Including the battery's internal resistance ($1 \, \Omega$), the total line current $I$ leaving the battery is:
$$I = \frac{V}{R_p + r} = \frac{9}{3 + 1} = \frac{9}{4} \mathrm{~A}$$
### Step 2: Find Current through the Main Branches
Using the current divider rule, the current $I_1$ flowing through the longer $12 \, \Omega$ branch is:
$$I_1 = I \times \frac{4}{12 + 4} = \frac{9}{4} \times \frac{4}{16} = \frac{9}{16} \mathrm{~A}$$
### Step 3: Find Potential Difference across the Diagonal
In a steady state, the capacitor acts as an open circuit. Let the diagonal link nodes be $A$ and $B$. The path contains two sides of the $12 \, \Omega$ branch (total resistance = $8 \, \Omega$):
$$V_C = V_A - V_B = I_1 \times 8 = \frac{9}{16} \times 8 = \frac{9}{2} \mathrm{~V}$$
### Step 4: Compute Stored Energy and Extract x
The stored energy $U$ is:
$$U = \frac{1}{2} \times (4 \, \mu \mathrm{F}) \times \left(\frac{9}{2}\right)^2 = \frac{1}{2} \times 4 \times \frac{81}{4} = \frac{81}{2} \, \mu \mathrm{J}$$
Comparing this directly with the expression $\frac{x}{2} \, \mu \mathrm{J}$:
$x = 81$
Therefore, the value of $x$ is $81$.
### Pattern Recognition
Remember that capacitors act as standard open circuits when reaching steady-state DC conditions. Simply calculate the node potentials across the bridge connection points using standard current distribution laws first, then execute the energy equation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Class 12 Physics: Electrostatic Potential and Capacitance
Q32 (2024)
The electric current through a wire varies with time as $I = I_0 + \beta t$, where $I_0 = 20 \mathrm{~A}$ and $\beta = 3 \mathrm{~A / s}$. The amount of electric charge crossed through a section of the wire in $20 \mathrm{~s}$ is:
- $80 \mathrm{~C}$
- $1000 \mathrm{~C}$
- $800 \mathrm{~C}$
- $1600 \mathrm{~C}$
### Related Formula
The relationship between current, charge, and time is given by:
$$I = \frac{dq}{dt} \implies q = \int I \, dt$$
### Core Logic
Given the current variation:
$$I = I_0 + \beta t$$
Substituting the given values, $I_0 = 20 \mathrm{~A}$ and $\beta = 3 \mathrm{~A/s}$:
$I = 20 + 3t$
Thus:
$$\frac{dq}{dt} = 20 + 3t$$
### Step 1: Integrate to Find Charge
To find the total charge crossing a section of wire from $t = 0$ to $t = 20 \mathrm{~s}$:
$$\int_{0}^{q} dq = \int_{0}^{20} (20 + 3t) \, dt$$
$$q = \left[ 20t + \frac{3t^2}{2} \right]_{0}^{20}$$
$$q = 20(20) + \frac{3(20)^2}{2}$$
$$q = 400 + \frac{3(400)}{2} = 400 + 600 = 1000 \mathrm{~C}$$
Therefore, the charge crossed is $1000 \mathrm{~C}$.
### Pattern Recognition
Whenever current is given as a function of time $I(t)$, the total charge is simply the area under the $I-t$ curve, which mathematically corresponds to the definite integral $\int_{t_1}^{t_2} I(t) \, dt$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q1740 (2024)
Two resistance of $100\Omega$ and $200\Omega$ are connected in series with a battery of $4 \mathrm{~V}$ and negligible internal resistance. A voltmeter is used to measure voltage across $100\Omega$ resistance, which gives reading as $1 \mathrm{~V}$. The resistance of voltmeter must be ________ $\Omega$.
### Related Formula
$$V = I R_{\text{eq}}$$
$$R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2}$$
### Core Logic
{{SOL_IMG1}}
The voltmeter has some internal resistance $R_v$ and is connected in parallel with the $100\Omega$ resistor.
The equivalent resistance of this parallel combination is $R_p = \frac{100 R_v}{100 + R_v}$.
This combination is in series with the $200\Omega$ resistor. The total voltage applied is $4 \mathrm{~V}$.
### Step 1: Set up Voltage Divider
The voltage across the parallel combination (the voltmeter reading) is $1 \mathrm{~V}$.
Therefore, the voltage across the $200\Omega$ resistor must be $4 \mathrm{~V} - 1 \mathrm{~V} = 3 \mathrm{~V}$.
Using the voltage divider rule (or equating currents since they are in series):
$$I = \frac{V_{200}}{200} = \frac{3}{200} \mathrm{~A}$$
### Step 2: Solve for Voltmeter Resistance
The current $I$ also flows through the parallel combination $R_p$:
$V_p = I R_p$
$$1 = \left(\frac{3}{200}\right) \left( \frac{100 R_v}{100 + R_v} \right)$$
$$200 (100 + R_v) = 300 R_v$$
$$20000 + 200 R_v = 300 R_v$$
$100 R_v = 20000$
$$R_v = 200 \Omega$$
### Pattern Recognition
If the voltage splits as $1\mathrm{V}$ to $3\mathrm{V}$, the resistances must be in a $1:3$ ratio. So $R_p = 200 / 3$. Equating $100 R_v / (100+R_v) = 200/3$ instantly gives $R_v = 200$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q48 (2024)
When a potential difference $V$ is applied across a wire of resistance $R$, it dissipates energy at a rate $W$. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:
- $1 / 4 \mathrm{W}$
- $1 / 2 \mathrm{W}$
- $2 \mathrm{~W}$
- $4 \mathrm{~W}$
### Related Formula
$$P = \frac{V^2}{R_{\text{eq}}}$$
### Core Logic
Initially, the power (rate of energy dissipation) is $W = \frac{V^2}{R}$.
When the wire is cut into two halves, each half will have a resistance of $\frac{R}{2}$ (since resistance is directly proportional to length).
### Step 1: Equivalent Resistance
When these two halves (each of $\frac{R}{2}$) are connected in parallel, their equivalent resistance $R_{\text{eq}}$ is:
$$R_{\text{eq}} = \frac{\left(\frac{R}{2}\right) \times \left(\frac{R}{2}\right)}{\frac{R}{2} + \frac{R}{2}} = \frac{R}{4}$$
### Step 2: New Power Dissipation
The new rate of energy dissipation $W'$ across the same potential difference $V$ is:
$$W' = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{\frac{R}{4}} = 4 \frac{V^2}{R}$$
$W' = 4W$
### Pattern Recognition
Cutting a resistor into $n$ equal parts and connecting them in parallel reduces the total resistance by a factor of $n^2$. Consequently, for a constant voltage supply, the power dissipated increases by a factor of $n^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q1822 (2024)
Two cells are connected in opposition as shown. Cell $E_1$ is of $8\mathrm{V}$ emf and $2\Omega$ internal resistance; the cell $E_2$ is of $2\mathrm{V}$ emf and $4\Omega$ internal resistance. The terminal potential difference of cell $E_2$ is:
{{IMG1}}
### Related Formula
$$I = \frac{E_{\text{net}}}{R_{\text{eq}}}$$
$$V = E - Ir \quad (\text{Discharging})$$
$$V = E + Ir \quad (\text{Charging})$$
### Core Logic
{{SOL_IMG1}}
Because the cells are in opposition, the net EMF drives current from the higher potential cell ($8\mathrm{V}$) to the lower potential cell ($2\mathrm{V}$). Thus, the $2\mathrm{V}$ cell acts as a load and undergoes charging.
### Step 1: Calculate Total Current
$$I = \frac{8 - 2}{2 + 4} = \frac{6}{6} = 1 \mathrm{~A}$$
### Step 2: Terminal Potential of Cell 2
Since cell $E_2$ is being charged, its terminal potential difference is:
$$V_2 = E_2 + I r_2$$
Applying Kirchhoff's rule across cell $E_2$ (from node C to B):
$$V_C - V_B = E_2 + I r_2 = 2 + (1)(4) = 6 \mathrm{~V}$$
### Pattern Recognition
When a smaller battery is forced backwards by a larger battery, it gets "charged". Consequently, its terminal voltage increases above its nominal EMF: $V = E + Ir$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
