A 16 \, Omega$16 \, \Omega$ wire is bend to form a square loop. A 9V battery with internal resistance 1 \, Omega$1 \, \Omega$ is connected across one of its sides. If a 4 \, mu mathrmF$4 \, \mu \mathrm{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be fracx2 \, mu mathrmJ$\frac{x}{2} \, \mu \mathrm{J}$. where x = $x = $ _________.
Numerical Answer Type:
Enter a numerical valueAnswer: 81 to 81+4 marks
Solution & Explanation
### Related Formula
Energy stored (U$U$) by a capacitor of capacitance C$C$ under steady state voltage V_C$V_C$:
U = frac12 C V_C^2$U = \frac{1}{2} C V_C^2$
### Core Logic
A wire of resistance 16 \, Omega$16 \, \Omega$ is bent into a square, so each of the 4 sides has a resistance of:
R_textside = frac164 = 4 \, Omega$R_{\text{side}} = \frac{16}{4} = 4 \, \Omega$
The battery is connected across one side. Let this side have resistance 4 \, Omega$4 \, \Omega$. The remaining three sides are connected in series, creating a parallel branch with a combined resistance of:
R_textseries = 4 + 4 + 4 = 12 \, Omega$R_{\text{series}} = 4 + 4 + 4 = 12 \, \Omega$Circuit network layout showing the square loop resistance divisions and diagonal capacitor connection for Q57
### Step 1: Calculate Equivalent Resistance and Circuit Current
The equivalent external resistance of the parallel loop branches is:
R_p = frac12 times 412 + 4 = frac4816 = 3 \, Omega$R_p = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 \, \Omega$
Including the battery's internal resistance (1 \, Omega$1 \, \Omega$), the total line current I$I$ leaving the battery is:
I = fracVR_p + r = frac93 + 1 = frac94 mathrm~A$I = \frac{V}{R_p + r} = \frac{9}{3 + 1} = \frac{9}{4} \mathrm{~A}$
### Step 2: Find Current through the Main Branches
Using the current divider rule, the current I_1$I_1$ flowing through the longer 12 \, Omega$12 \, \Omega$ branch is:
I_1 = I times frac412 + 4 = frac94 times frac416 = frac916 mathrm~A$I_1 = I \times \frac{4}{12 + 4} = \frac{9}{4} \times \frac{4}{16} = \frac{9}{16} \mathrm{~A}$
### 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$A$ and B$B$. The path contains two sides of the 12 \, Omega$12 \, \Omega$ branch (total resistance = 8 \, Omega$8 \, \Omega$):
V_C = V_A - V_B = I_1 times 8 = frac916 times 8 = frac92 mathrm~V$V_C = V_A - V_B = I_1 \times 8 = \frac{9}{16} \times 8 = \frac{9}{2} \mathrm{~V}$
### Step 4: Compute Stored Energy and Extract x
The stored energy U$U$ is:
U = frac12 times (4 \, mu mathrmF) times left(frac92right)^2 = frac12 times 4 times frac814 = frac812 \, mu mathrmJ$U = \frac{1}{2} \times (4 \, \mu \mathrm{F}) \times \left(\frac{9}{2}\right)^2 = \frac{1}{2} \times 4 \times \frac{81}{4} = \frac{81}{2} \, \mu \mathrm{J}$
Comparing this directly with the expression fracx2 \, mu mathrmJ$\frac{x}{2} \, \mu \mathrm{J}$:
x = 81$x = 81$
Therefore, the value of x$x$ is 81$81$.
### 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
Keywords:#energy stored by the capacitor will be x/2#JEE Main 2024 Morning Q57#Current Electricity JEE Main 2024#Kirchhoff's Laws and RC Circuits
More Current Electricity Previous-Year Questions — Page 3
Q13jee_main_2025_04_april_morningElectric Current and Charge Flow
Current passing through a wire as function of time is given as I(t)=0.02t+0.01mathrm~A$I(t)=0.02t+0.01\mathrm{~A}$. The charge that will flow through the wire from t=1mathrm~s$t=1\mathrm{~s}$ to t=2mathrm~s$t=2\mathrm{~s}$ is:
### Core Logic
The total resistance of the continuous uniform wire loop is 9Omega$9\Omega$. When bent into an equilateral triangle, it is split into three equal length sections. The resistance of each individual side is:
R_textside = frac9Omega3 = 3Omega$R_{\text{side}} = \frac{9\Omega}{3} = 3\Omega$
### Step 1: Calculating Equivalent Series and Parallel Resistance
As shown in the circuit diagrams Combination of Resistors diagram for Q23 - JEE Main 2025 Morning and Combination of Resistors diagram for Q23 - JEE Main 2025 Morning, measuring across any two vertices means one branch contains a single side resistor (3Omega$3\Omega$), while the other branch contains the remaining two sides connected in series :
R_textseries = 3Omega + 3Omega = 6Omega$R_{\text{series}} = 3\Omega + 3\Omega = 6\Omega$
Now, calculate the parallel equivalent between these two branches:
R_texteq = fracR_textside times R_textseriesR_textside + R_textseries = frac3 times 63 + 6 = frac189 = 2Omega$R_{\text{eq}} = \frac{R_{\text{side}} \times R_{\text{series}}}{R_{\text{side}} + R_{\text{series}}} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\Omega$
### Pattern Recognition
For a closed uniform loop with N$N$ equal sides, the parallel resistance measured across adjacent corners always simplifies to fracN-1N^2 cdot R_texttotal$\frac{N-1}{N^2} \cdot R_{\text{total}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q24jee_main_2025_28_jan_eveningWheatstone Bridge
The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be ________ A.
A bridge resistor network supplied by a 40V DC voltage source terminal layout.
Numerical Answer.Answer: 2
Solution
### Related Formula
For a balanced Wheatstone bridge network, if the potentials at opposite nodes are equal (V_A = V_B$V_A = V_B$), no current flows through the central branch. The resistance arms satisfy the balance ratio:
fracR_1R_2 = fracR_3R_4$\frac{R_1}{R_2} = \frac{R_3}{R_4}$
Total current from the source is calculated using Ohm's law:
I = fracV_textsourceR_textequivalent$I = \frac{V_{\text{source}}}{R_{\text{equivalent}}}$
### Core Logic
Given conditions :
V_A = V_B implies textBalanced condition$V_A = V_B \implies \text{Balanced condition}$
From the circuit network diagram [cite: 817, 818, 826, 828]:
* Left-top arm = 10 \ Omega$= 10 \ \Omega$
* Left-bottom arm = R$= R$
* Right-top arm = 20 \ Omega$= 20 \ \Omega$
* Right-bottom arm = 40 \ Omega$= 40 \ \Omega$
Apply the balance condition to solve for unknown resistor R$R$ :
frac10R = frac2040 implies frac10R = frac12 implies R = 20 \ Omega quad text[cite: 837, 838]$\frac{10}{R} = \frac{20}{40} \implies \frac{10}{R} = \frac{1}{2} \implies R = 20 \ \Omega \quad \text{[cite: 837, 838]}$
Now restructure the equivalent network :
Since the central 30 \ Omega$30 \ \Omega$ resistor branch carries zero current, it can be removed from the calculation [cite: 820, 834].
* Top \parallel branch: 10 \ Omega + 20 \ Omega = 30 \ Omega$10 \ \Omega + 20 \ \Omega = 30 \ \Omega$
* Bottom \parallel branch: 20 \ Omega + 40 \ Omega = 60 \ Omega$20 \ \Omega + 40 \ \Omega = 60 \ \Omega$
Calculate total equivalent resistance R_texteq$R_{\text{eq}}$:
R_texteq = frac30 times 6030 + 60 = frac180090 = 20 \ Omega$R_{\text{eq}} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20 \ \Omega$
Calculate total current I$I$ drawn from the 40textV$40\text{V}$ source:
I = frac40 text V20 \ Omega = 2 text A$I = \frac{40 \text{ V}}{20 \ \Omega} = 2 \text{ A}$
### Step 1: Circuit Solution
The balanced bridge network layout with branch currents is shown below:
A bridge resistor network supplied by a 40V DC voltage source terminal layout.
### Pattern Recognition
When a question states that two nodes are at equal potential (V_A = V_B$V_A = V_B$), immediately identify it as a balanced Wheatstone bridge. This allows you to remove the central branch and simplify the circuit into basic series-\parallel resistor combinations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q34jee_main_2024_01_february_morningCells and EMF
The reading in the ideal voltmeter (V) shown in the given circuit diagram is:
A schematic showing a symmetric loop containing multiple identical 5V sources with 0.2 Ohm internal resistances coupled across an ideal voltmeter terminal.
A.5mathrm~V$5\mathrm{~V}$
B.10mathrm~V$10\mathrm{~V}$
C.0mathrm~V$0\mathrm{~V}$
D.3mathrm~V$3\mathrm{~V}$
Solution
### Related Formula
Total effective loop EMF and internal resistance:
I = fracE_textnetr_textnet$I = \frac{E_{\text{net}}}{r_{\text{net}}}$
Terminal potential difference across a discharging cell:
V = E - Ir$V = E - Ir$
### Core Logic
The loop has a set of 8 cells all aiding the same current flow sequence direction.
E_texteq = 8 times 5 = 40mathrm~V$E_{\text{eq}} = 8 \times 5 = 40\mathrm{~V}$r_texteq = 8 times 0.2 = 1.6mathrm~Omega$r_{\text{eq}} = 8 \times 0.2 = 1.6\mathrm{~\Omega}$
The circulating loop current is:
I = frac40mathrm~V1.6mathrm~Omega = 25mathrm~A$I = \frac{40\mathrm{~V}}{1.6\mathrm{~\Omega}} = 25\mathrm{~A}$
### Step 1: Terminal Voltage Calculation
The ideal voltmeter measures the potential drop across the localized parallel branch configuration:
V = E - Ir = 5 - (25 times 0.2) = 5 - 5 = 0mathrm~V$V = E - Ir = 5 - (25 \times 0.2) = 5 - 5 = 0\mathrm{~V}$
### Pattern Recognition
A closed loop composed entirely of identical series active cells short circuits itself perfectly relative to the localized node drop points, reducing the net external voltage drop to zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q59jee_main_2024_01_february_morningElectric Charge and Current
The current in a conductor is expressed as I = 3t^2 + 4t^3$I = 3t^2 + 4t^3$, where I$I$ is in Ampere and t$t$ is in second. The amount of electric charge that flows through a section of the conductor during t = 1mathrm~s$t = 1\mathrm{~s}$ to t = 2mathrm~s$t = 2\mathrm{~s}$ is _______ mathrmC$\mathrm{C}$.
Numerical Answer.Answer: 22 to 22
Solution
### Related Formula
Relationship between charge and time-varying current:
I = fracdqdt implies q = int_t_1^t_2 I \, dt$I = \frac{dq}{dt} \implies q = \int_{t_1}^{t_2} I \, dt$
### Core Logic
Set up the definite integral using the given bounds t=1mathrm~s$t=1\mathrm{~s}$ to t=2mathrm~s$t=2\mathrm{~s}$:
q = int_1^2 (3t^2 + 4t^3) \, dt$q = \int_{1}^{2} (3t^2 + 4t^3) \, dt$
Perform integration term-by-term:
q = left[ frac3t^33 + frac4t^44 right]_1^2 = left[ t^3 + t^4 right]_1^2$q = \left[ \frac{3t^3}{3} + \frac{4t^4}{4} \right]_1^2 = \left[ t^3 + t^4 \right]_1^2$
### Step 1: Evaluate Definite Bounds
Substitute upper and lower limits:
q = (2^3 + 2^4) - (1^3 + 1^4)$q = (2^3 + 2^4) - (1^3 + 1^4)$q = (8 + 16) - (1 + 1) = 24 - 2 = 22mathrm~C$q = (8 + 16) - (1 + 1) = 24 - 2 = 22\mathrm{~C}$
### Pattern Recognition
Simple polynomial integration. Always evaluate both boundary points explicitly to avoid dropped terms from the lower bound.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
More Current Electricity Questions — jee_main_2024_29_jan_morning
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