### Related Formula q = int_t_1^t_2 I(t) \, dt$q = \int_{t_1}^{t_2} I(t) \, dt$ ### Core Logic Given trace: I(t) = 0.02t + 0.01$I(t) = 0.02t + 0.01$ Limits: t_1 = 1mathrm~s$t_1 = 1\mathrm{~s}$, t_2 = 2mathrm~s$t_2 = 2\mathrm{~s}$ ### Step 1: Perform Definitive Integration q = int_1^2 (0.02t + 0.01) \, dt$q = \int_{1}^{2} (0.02t + 0.01) \, dt$ q = left[ 0.02fract^22 + 0.01t ight]_1^2 = left[ 0.01t^2 + 0.01t ight]_1^2$q = \left[ 0.02\frac{t^2}{2} + 0.01t ight]_{1}^{2} = \left[ 0.01t^2 + 0.01t ight]_{1}^{2}$ q = left[ 0.01(2)^2 + 0.01(2) ight] - left[ 0.01(1)^2 + 0.01(1) ight]$q = \left[ 0.01(2)^2 + 0.01(2) ight] - \left[ 0.01(1)^2 + 0.01(1) ight]$ q = [0.04 + 0.02] - [0.01 + 0.01] = 0.06 - 0.02 = 0.04mathrm~C$q = [0.04 + 0.02] - [0.01 + 0.01] = 0.06 - 0.02 = 0.04\mathrm{~C}$ Hence, the total charge integration yields 0.04mathrm~C$0.04\mathrm{~C}$. ### Pattern Recognition Definite calculus integrations over a linear function can also be checked visually via calculating trapezoidal graph spaces under the Itext-t$I\text{-}t$ curve line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### 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

and

, 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

### 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:

### 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

### 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

### 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