### Related Formula The relationship between current, charge, and time is given by: I = fracdqdt implies q = int I \, dt$I = \frac{dq}{dt} \implies q = \int I \, dt$ ### Core Logic Given the current variation: I = I_0 + beta t$I = I_0 + \beta t$ Substituting the given values, I_0 = 20 mathrm~A$I_0 = 20 \mathrm{~A}$ and beta = 3 mathrm~A/s$\beta = 3 \mathrm{~A/s}$: I = 20 + 3t$I = 20 + 3t$ Thus: fracdqdt = 20 + 3t$\frac{dq}{dt} = 20 + 3t$ ### Step 1: Integrate to Find Charge To find the total charge crossing a section of wire from t = 0$t = 0$ to t = 20 mathrm~s$t = 20 \mathrm{~s}$: int_0^q dq = int_0^20 (20 + 3t) \, dt$\int_{0}^{q} dq = \int_{0}^{20} (20 + 3t) \, dt$ q = left[ 20t + frac3t^22 right]_0^20$q = \left[ 20t + \frac{3t^2}{2} \right]_{0}^{20}$ q = 20(20) + frac3(20)^22$q = 20(20) + \frac{3(20)^2}{2}$ q = 400 + frac3(400)2 = 400 + 600 = 1000 mathrm~C$q = 400 + \frac{3(400)}{2} = 400 + 600 = 1000 \mathrm{~C}$ Therefore, the charge crossed is 1000 mathrm~C$1000 \mathrm{~C}$. ### Pattern Recognition Whenever current is given as a function of time I(t)$I(t)$, the total charge is simply the area under the I-t$I-t$ curve, which mathematically corresponds to the definite integral int_t_1^t_2 I(t) \, dt$\int_{t_1}^{t_2} I(t) \, dt$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### 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

### Related Formula P = fracV^2R_texteq$P = \frac{V^2}{R_{\text{eq}}}$ ### Core Logic Initially, the power (rate of energy dissipation) is W = fracV^2R$W = \frac{V^2}{R}$. When the wire is cut into two halves, each half will have a resistance of fracR2$\frac{R}{2}$ (since resistance is directly proportional to length). ### Step 1: Equivalent Resistance When these two halves (each of fracR2$\frac{R}{2}$) are connected in parallel, their equivalent resistance R_texteq$R_{\text{eq}}$ is: R_texteq = fracleft(fracR2right) times left(fracR2right)fracR2 + fracR2 = fracR4$R_{\text{eq}} = \frac{\left(\frac{R}{2}\right) \times \left(\frac{R}{2}\right)}{\frac{R}{2} + \frac{R}{2}} = \frac{R}{4}$ ### Step 2: New Power Dissipation The new rate of energy dissipation W'$W'$ across the same potential difference V$V$ is: W' = fracV^2R_texteq = fracV^2fracR4 = 4 fracV^2R$W' = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{\frac{R}{4}} = 4 \frac{V^2}{R}$ W' = 4W$W' = 4W$ ### Pattern Recognition Cutting a resistor into n$n$ equal parts and connecting them in parallel reduces the total resistance by a factor of n^2$n^2$. Consequently, for a constant voltage supply, the power dissipated increases by a factor of n^2$n^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula V = I R_texteq$V = I R_{\text{eq}}$ R_textparallel = fracR_1 R_2R_1 + R_2$R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2}$ ### Core Logic

The voltmeter has some internal resistance R_v$R_v$ and is connected in parallel with the 100Omega$100\Omega$ resistor. The equivalent resistance of this parallel combination is R_p = frac100 R_v100 + R_v$R_p = \frac{100 R_v}{100 + R_v}$. This combination is in series with the 200Omega$200\Omega$ resistor. The total voltage applied is 4 mathrm~V$4 \mathrm{~V}$. ### Step 1: Set up Voltage Divider The voltage across the parallel combination (the voltmeter reading) is 1 mathrm~V$1 \mathrm{~V}$. Therefore, the voltage across the 200Omega$200\Omega$ resistor must be 4 mathrm~V - 1 mathrm~V = 3 mathrm~V$4 \mathrm{~V} - 1 \mathrm{~V} = 3 \mathrm{~V}$. Using the voltage divider rule (or equating currents since they are in series): I = fracV_200200 = frac3200 mathrm~A$I = \frac{V_{200}}{200} = \frac{3}{200} \mathrm{~A}$ ### Step 2: Solve for Voltmeter Resistance The current I$I$ also flows through the parallel combination R_p$R_p$: V_p = I R_p$V_p = I R_p$ 1 = left(frac3200right) left( frac100 R_v100 + R_v right)$1 = \left(\frac{3}{200}\right) \left( \frac{100 R_v}{100 + R_v} \right)$ 200 (100 + R_v) = 300 R_v$200 (100 + R_v) = 300 R_v$ 20000 + 200 R_v = 300 R_v$20000 + 200 R_v = 300 R_v$ 100 R_v = 20000$100 R_v = 20000$ R_v = 200 Omega$R_v = 200 \Omega$ ### Pattern Recognition If the voltage splits as 1mathrmV$1\mathrm{V}$ to 3mathrmV$3\mathrm{V}$, the resistances must be in a 1:3$1:3$ ratio. So R_p = 200 / 3$R_p = 200 / 3$. Equating 100 R_v / (100+R_v) = 200/3$100 R_v / (100+R_v) = 200/3$ instantly gives R_v = 200$R_v = 200$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula V = IR$V = IR$ R_texteq = R_1 + R_2 + dots + R_n quad (textfor series)$R_{\text{eq}} = R_1 + R_2 + \dots + R_n \quad (\text{for series})$ ### Core Logic Observe the circuit diagram. The total equivalent resistance R_texteq$R_{\text{eq}}$ 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$4000 \,\Omega$ (comprising the 3.3mathrmkOmega$3.3\mathrm{k}\Omega$ resistor and seven 100\,Omega$100\,\Omega$ resistors). R_texteq = 4000 \,Omega$R_{\text{eq}} = 4000 \,\Omega$ The total voltage applied across the network is 4 mathrm~V$4 \mathrm{~V}$. i = fracVR_texteq = frac44000 = frac11000 mathrm~A$i = \frac{V}{R_{\text{eq}}} = \frac{4}{4000} = \frac{1}{1000} \mathrm{~A}$ ### Step 2: Calculate Output Voltage The output voltage V_0$V_0$ is tapped across five 100 \,Omega$100 \,\Omega$ resistors. R_textout = 5 times 100 \,Omega = 500 \,Omega$R_{\text{out}} = 5 \times 100 \,\Omega = 500 \,\Omega$ Thus, the output voltage is: V_0 = i cdot R_textout = left(frac11000right) times 500 = 0.5 mathrm~V$V_0 = i \cdot R_{\text{out}} = \left(\frac{1}{1000}\right) \times 500 = 0.5 \mathrm{~V}$ ### 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_texttap / R_texttotal)$V_0 = V_{in} \times (R_{\text{tap}} / R_{\text{total}})$. Standard DC circuit division. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity