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 4
Q46jee_main_2024_29_january_eveningSeries and Parallel Combination of Resistors
In the given circuit, the current in resistance R_3$R_3$ is:
The diagram displays a circuit consisting of series-parallel combinations of R1, R2, R3, and R4 with a 10V voltage source.
A.1text A$1\text{ A}$
B.1.5text A$1.5\text{ A}$
C.2text A$2\text{ A}$
D.2.5text A$2.5\text{ A}$
Solution
### Related Formula
For parallel resistors:
R_textparallel = fracR_a R_bR_a + R_b$R_{\text{parallel}} = \frac{R_a R_b}{R_a + R_b}$
Total equivalent resistance in series:
R_texteq = R_1 + R_textparallel + R_4$R_{\text{eq}} = R_1 + R_{\text{parallel}} + R_4$
Total current from source:
I = fracVR_texteq$I = \frac{V}{R_{\text{eq}}}$
### Core Logic
Analyzing the circuit network:
* R_1 = 2\ Omega$R_1 = 2\ \Omega$
* R_2 = 4\ Omega$R_2 = 4\ \Omega$ and R_3 = 4\ Omega$R_3 = 4\ \Omega$ are in parallel.
* R_4 = 1\ Omega$R_4 = 1\ \Omega$
Calculate the equivalent resistance of the parallel combination:
R_textparallel = fracR_2 times R_3R_2 + R_3 = frac4 times 44 + 4 = 2\ Omega$R_{\text{parallel}} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{4 \times 4}{4 + 4} = 2\ \Omega$
### Step 1: Calculate Total Equivalent Resistance and Current
Total equivalent resistance is:
R_texteq = R_1 + R_textparallel + R_4 = 2 + 2 + 1 = 5\ Omega$R_{\text{eq}} = R_1 + R_{\text{parallel}} + R_4 = 2 + 2 + 1 = 5\ \Omega$
Total circuit current is:
I = fracVR_texteq = frac10text V5\ Omega = 2text A$I = \frac{V}{R_{\text{eq}}} = \frac{10\text{ V}}{5\ \Omega} = 2\text{ A}$
### Step 2: Determine Current in R3
The total current of 2text A$2\text{ A}$ enters the parallel branch of R_2$R_2$ and R_3$R_3$. Since R_2 = R_3 = 4\ Omega$R_2 = R_3 = 4\ \Omega$, the current divides equally between them:
I_R_3 = I times fracR_2R_2 + R_3 = 2 times frac48 = 1text A$I_{R_3} = I \times \frac{R_2}{R_2 + R_3} = 2 \times \frac{4}{8} = 1\text{ A}$The diagram displays a circuit consisting of series-parallel combinations of R1, R2, R3, and R4 with a 10V voltage source.
### Pattern Recognition
Equal parallel resistors split current exactly down the middle. Once total current is found to be 2text A$2\text{ A}$, the parallel branches share it as 1text A$1\text{ A}$ each without further math.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q58jee_main_2024_29_january_eveningKirchhoff's Laws and Mesh Analysis
In the given circuit, the current flowing through the resistance 20\ Omega$20\ \Omega$ is 0.3text A$0.3\text{ A}$, while the ammeter reads 0.9text A$0.9\text{ A}$. The value of R_1$R_1$ is ________ Omega$\Omega$.
The diagram displays a circuit consisting of three parallel branches containing R1, a 20 Ohm resistor, and a 15 Ohm resistor, with an ammeter in series.
Numerical Answer.Answer: 30 to 30
Solution
### Related Formula
For parallel branches, the potential difference V$V$ across each branch is identical:
V = I_i R_i$V = I_i R_i$
According to Kirchhoff's Current Law, the total current I_texttotal$I_{\text{total}}$ is the sum of currents in all parallel branches:
I_texttotal = i_1 + i_2 + i_3$I_{\text{total}} = i_1 + i_2 + i_3$
### Core Logic
Analyzing the circuit diagram:
* Branch 1: Current i_1$i_1$ through 20\ Omega$20\ \Omega$ is 0.3text A$0.3\text{ A}$.
* Branch 2: Contains 15\ Omega$15\ \Omega$ resistor with current i_2$i_2$.
* Branch 3: Contains resistor R_1$R_1$ with current i_3$i_3$.
Since the branches are in parallel, they have the same potential difference V_AB$V_{AB}$:
V_AB = i_1 times 20\ Omega = 0.3text A times 20\ Omega = 6text V$V_{AB} = i_1 \times 20\ \Omega = 0.3\text{ A} \times 20\ \Omega = 6\text{ V}$
### Step 1: Calculate Currents
Current through the second branch (15\ Omega$15\ \Omega$ resistor) is:
i_2 = fracV_AB15\ Omega = frac6text V15\ Omega = 0.4text A$i_2 = \frac{V_{AB}}{15\ \Omega} = \frac{6\text{ V}}{15\ \Omega} = 0.4\text{ A}$
Total current read by the ammeter is 0.9text A$0.9\text{ A}$. Thus:
i_1 + i_2 + i_3 = 0.9text A$i_1 + i_2 + i_3 = 0.9\text{ A}$0.3text A + 0.4text A + i_3 = 0.9text A$0.3\text{ A} + 0.4\text{ A} + i_3 = 0.9\text{ A}$0.7text A + i_3 = 0.9text A implies i_3 = 0.2text 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$R_1$:
i_3 times R_1 = V_AB$i_3 \times R_1 = V_{AB}$(0.2text A) times R_1 = 6text V$(0.2\text{ A}) \times R_1 = 6\text{ V}$R_1 = frac60.2 = 30\ Omega$R_1 = \frac{6}{0.2} = 30\ \Omega$The diagram displays a circuit consisting of three parallel branches containing R1, a 20 Ohm resistor, and a 15 Ohm resistor, with an ammeter in series.
### Pattern Recognition
In parallel networks, finding the branch voltage is always the primary step. Once V = 6text V$V = 6\text{ V}$ 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
Q48jee_main_2024_27_jan_morningMeter Bridge and Resistivity
A wire of length 10text cm$10\text{ cm}$ and radius sqrt7 times 10^-4text m$\sqrt{7} \times 10^{-4}\text{ m}$ is connected across the right gap of a meter bridge. When a resistance of 4.5\ Omega$4.5\ \Omega$ is connected on the left gap by using a resistance box, the balance length is found to be at 60text cm$60\text{ cm}$ from the left end.
If the resistivity of the wire is R times 10^-7\ Omegatextm$R \times 10^{-7}\ \Omega\text{m}$, then the value of R$R$ is:
A.63$63$
B.70$70$
C.66$66$
D.35$35$
Solution
### Related Formula
From the balance condition of the meter bridge:
fracX_textleftl = fracX_textright100 - l$\frac{X_{\text{left}}}{l} = \frac{X_{\text{right}}}{100 - l}$
Resistance formula:
X = rho fracl_wA = fracrho l_wpi r^2$X = \rho \frac{l_w}{A} = \frac{\rho l_w}{\pi r^2}$
### Core Logic
First, evaluate the unknown resistance X_textright$X_{\text{right}}$ in the right gap:
frac4.560 = fracX_textright40 implies X_textright = frac4.5 times 4060 = 3\ Omega$\frac{4.5}{60} = \frac{X_{\text{right}}}{40} \implies X_{\text{right}} = \frac{4.5 \times 40}{60} = 3\ \Omega$
### Step 1: Calculate Resistivity Value
Now map the resistance parameters (l_w = 10text cm = 0.1text m$l_w = 10\text{ cm} = 0.1\text{ m}$, r = sqrt7 times 10^-4text m$r = \sqrt{7} \times 10^{-4}\text{ m}$):
3 = rho frac0.1frac227 times (sqrt7 times 10^-4)^2$3 = \rho \frac{0.1}{\frac{22}{7} \times (\sqrt{7} \times 10^{-4})^2}$3 = rho frac0.1frac227 times 7 times 10^-8$3 = \rho \frac{0.1}{\frac{22}{7} \times 7 \times 10^{-8}}$3 = rho frac0.122 times 10^-8$3 = \rho \frac{0.1}{22 \times 10^{-8}}$rho = frac3 times 22 times 10^-80.1 = 66 times 10^-7\ Omegatextm$\rho = \frac{3 \times 22 \times 10^{-8}}{0.1} = 66 \times 10^{-7}\ \Omega\text{m}$
### Step 2: Compare to find R
Given rho = R times 10^-7$\rho = R \times 10^{-7}$, comparing coefficients gives:
R = 66$R = 66$
### Pattern Recognition
Meter bridge balance simplifies directly to simple scalar component checks. The sqrt7$\sqrt{7}$ term perfectly neutralizes the fractional frac227$\frac{22}{7}$ constant in circular area profiles.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q49jee_main_2024_27_jan_morningCombination of Resistors
A wire of resistance R$R$ and length L$L$ is cut into 5$5$ equal parts. If these parts are joined parallely, then the resultant resistance will be:
A.frac125R$\frac{1}{25}R$
B.frac15R$\frac{1}{5}R$
C.25R$25R$
D.5R$5R$
Solution
### Core Logic
Resistance is directly proportional to length (R propto L$R \propto L$). Cutting the wire into 5 equal pieces reduces the resistance of each segment to:
R' = fracR5$R' = \frac{R}{5}$
### Step 1: Compute parallel value
Connecting 5 identical resistors R'$R'$ in parallel gives a total equivalent resistance of:
R_texteq = fracR'5 = fracR/55 = fracR25$R_{\text{eq}} = \frac{R'}{5} = \frac{R/5}{5} = \frac{R}{25}$
### Pattern Recognition
Cutting an item into N$N$ components and grouping them in parallel scales the overall baseline systemic value down cleanly by a factor of N^2$N^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q59jee_main_2024_27_jan_morningCapacitor in DC Circuit
The charge accumulated on the capacitor connected in the following circuit is ______ mutextC$\mu\text{C}$. (Given C = 150\ mutextF$C = 150\ \mu\text{F}$).
The layout shows an integrated network of resistors labeled R1 to R4 powered by a 10V DC source loop with a branch housing a 150 uF capacitor element bridging distinct structural potential nodes.
Numerical Answer.Answer: 400 to 400
Solution
### Related Formula
Q = C cdot Delta V_C$Q = C \cdot \Delta V_C$
### 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 + frac103(1) - 6(1) = V_B$V_A + \frac{10}{3}(1) - 6(1) = V_B$
### Step 1: Compute node potential difference
V_A - V_B = 6 - frac103 = frac83text V$V_A - V_B = 6 - \frac{10}{3} = \frac{8}{3}\text{ V}$
### Step 2: Evaluate accumulated charge
Q = C(V_A - V_B) = 150\ mutextF times frac83text V = 50 times 8 = 400\ mutextC$Q = C(V_A - V_B) = 150\ \mu\text{F} \times \frac{8}{3}\text{ V} = 50 \times 8 = 400\ \mu\text{C}$
### 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
More Current Electricity Questions — jee_main_2024_29_jan_morning
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