### 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$R_{\text{series}} = 4 + 4 + 4 = 12 \, \Omega$

### 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