The battery of a mobile phone is rated as 4.2mathrm~V$4.2\mathrm{~V}$, 5800mathrm~mAh$5800\mathrm{~mAh}$. How much energy is stored in it when fully charged?

### 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

### 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$$I_{R_3} = I \times \frac{R_2}{R_2 + R_3} = 2 \times \frac{4}{8} = 1\text{ A}$$

### 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

### 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$$R_1 = \frac{6}{0.2} = 30\ \Omega$$

### 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

### 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

### 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