### Related Formula E = V cdot Q$E = V \cdot Q$ Q = I cdot t$Q = I \cdot t$ ### Core Logic The capacity rating 5800mathrm~mAh$5800\mathrm{~mAh}$ represents total electric charge Q$Q$ stored: Q = 5800mathrm~mA times 1mathrm~hour = (5800 times 10^-3mathrm~A) times 3600mathrm~s = 20880mathrm~C$Q = 5800\mathrm{~mA} \times 1\mathrm{~hour} = (5800 \times 10^{-3}\mathrm{~A}) \times 3600\mathrm{~s} = 20880\mathrm{~C}$ Given the voltage is V = 4.2mathrm~V$V = 4.2\mathrm{~V}$, the energy stored in Joules is: E = V cdot Q = 4.2 times 20880 = 87696mathrm~J = 87.696mathrm~kJ approx 87.7mathrm~kJ$E = V \cdot Q = 4.2 \times 20880 = 87696\mathrm{~J} = 87.696\mathrm{~kJ} \approx 87.7\mathrm{~kJ}$ ### Step 1: Final Conclusion The electrical energy stored in the fully charged battery is 87.7mathrm~kJ$87.7\mathrm{~kJ}$. ### Pattern Recognition Always remember: textEnergy (in Joules) = textVoltage (V) times textCapacity (Ah) times 3600$\text{Energy (in Joules)} = \text{Voltage (V)} \times \text{Capacity (Ah)} \times 3600$. This directly transforms electrical rating units to standard SI energy units. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula - Series cell configuration: varepsilon_texteq = varepsilon_1 + varepsilon_2, quad r_texteq = r_1 + r_2$\varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2, \quad r_{\text{eq}} = r_1 + r_2$ - Parallel cell configuration (for unequal cells in parallel): varepsilon_texteq = fracfracvarepsilon_1r_1 + fracvarepsilon_2r_2frac1r_1 + frac1r_2, quad frac1r_texteq = frac1r_1 + frac1r_2$\varepsilon_{\text{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}, \quad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}$ - Circuit current: I = fracvarepsilon_texteqr_texteq + R$I = \frac{\varepsilon_{\text{eq}}}{r_{\text{eq}} + R}$ ### Core Logic Given parameters: - Cell 1: $ ### Core Logic Given parameters: - Cell 1: $\varepsilon_1 = 1\mathrm{~V}, $, $r_1 = 2\Omega - Cell 2: $ - Cell 2: $\varepsilon_2 = 2\mathrm{~V}, $, $r_2 = 1\Omega - External resistance $ - External resistance $R = 6\Omega ### Step 1: Calculate $ ### Step 1: Calculate $I_1 (Series Configuration)

$ (Series Configuration)

$varepsilon_texteq = 1 + 2 = 3mathrm~V$\varepsilon_{\text{eq}} = 1 + 2 = 3\mathrm{~V}$ $ $r_texteq = 2 + 1 = 3Omega$r_{\text{eq}} = 2 + 1 = 3\Omega$ $ $I_1 = frac33 + 6 = frac39 = frac13mathrm~A$I_1 = \frac{3}{3 + 6} = \frac{3}{9} = \frac{1}{3}\mathrm{~A}$ ### Step 2: Calculate $I_2$ (Parallel Configuration)

$ ### Step 2: Calculate $I_2$ (Parallel Configuration)

$\varepsilon_{\text{eq}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{0.5 + 2}{1.5} = \frac{2.5}{1.5} = \frac{5}{3}\mathrm{~V} $ $r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\Omega $ $I_2 = \frac{\varepsilon_{\text{eq}}}{r_{\text{eq}} + R} = \frac{\frac{5}{3}}{\frac{2}{3} + 6} = \frac{\frac{5}{3}}{\frac{20}{3}} = \frac{5}{20} = \frac{1}{4}\mathrm{~A}$ ### Step 3: Find ratio and evaluate $x$ $$ ### Step 3: Find ratio and evaluate $x$ $\frac{I_1}{I_2} = \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{4}{3}$ Comparing with $ Comparing with $\frac{x}{3}: $: $x = 4 ### Pattern Recognition For cells connected in parallel, the formula for equivalent EMF $ ### Pattern Recognition For cells connected in parallel, the formula for equivalent EMF $\varepsilon_{\text{eq}}$ can be viewed as a weighted average. Connecting cells in series maximizes EMF but increases internal resistance, while parallel configuration limits EMF to an intermediate value while reducing equivalent internal resistance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula For a wire of total resistance R$R$ divided into N$N$ equal segments, the resistance of each segment r$r$ is: r = fracRN$r = \frac{R}{N}$ For a balanced Wheatstone bridge with resistors of resistance r$r$, the central arm can be neglected because no current flows through it. ### Core Logic The triangular pyramid has 6$6$ segments of equal length. Since the total resistance of the wire is R$R$: r = fracR6$r = \frac{R}{6}$ Let the four vertices of the pyramid be A$A$, B$B$, C$C$, and D$D$. Terminals are at A$A$ and B$B$. The segments are: - AB$AB$ (direct path between terminals, resistance r$r$) - AC$AC$, BC$BC$, AD$AD$, BD$BD$ (forming a closed quadrilateral network between A$A$ and B$B$ with bridge arm CD$CD$) - CD$CD$ (bridge arm connecting the midpoints, resistance r$r$) ### Step 1: Simplify the Network By symmetry, the potentials at C$C$ and D$D$ are equal when a voltage is applied across A$A$ and B$B$. Thus, the bridge is balanced, and no current flows through the segment CD$CD$. We can remove segment CD$CD$ from the calculations: - The path A to C to B$A \to C \to B$ consists of two resistors in series: r + r = 2r$r + r = 2r$. - The path A to D to B$A \to D \to B$ also consists of two resistors in series: r + r = 2r$r + r = 2r$. - The direct path A to B$A \to B$ has a single resistor r$r$. These three parallel branches are connected between A$A$ and B$B$. ### Step 2: Calculate Equivalent Resistance The equivalent resistance R_AB$R_{AB}$ is: frac1R_AB = frac12r + frac12r + frac1r = frac1r + frac1r = frac2r$\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$ R_AB = fracr2$R_{AB} = \frac{r}{2}$ Substitute r = fracR6$r = \frac{R}{6}$: R_AB = fracR/62 = fracR12$R_{AB} = \frac{R/6}{2} = \frac{R}{12}$ Comparing with R_AB = fracRn$R_{AB} = \frac{R}{n}$, we find n = 12$n = 12$. ### Pattern Recognition Sees: Resistor network formed by a 3D pyramid (6 identical edges, 4 nodes). Shortcut: A 6-resistor regular tetrahedron has an equivalent resistance of r/2$r/2$ across any two vertices. Since the total wire resistance is R$R$ and it's cut into 6 pieces, r = R/6 implies R_texteq = R/12$r = R/6 \implies R_{\text{eq}} = R/12$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula V = S cdot Delta T$V = S \cdot \Delta T$ where, V$V$ = Thermoelectric voltage (Seebeck voltage) S$S$ = Seebeck coefficient Delta T$\Delta T$ = Temperature difference ### Core Logic To maximize the efficiency of a thermoelectric device harvesting heat energy, two conditions must be fulfilled: 1. **Low thermal conductivity**: This ensures that the temperature gradient (Delta T$\Delta T$) across the material is maintained and heat does not rapidly flow from the hot side to the cold side. 2. **High electrical conductivity**: This minimizes internal Joule heating losses (I^2R$I^2R$) when electrical current is drawn from the material. Therefore, the material should possess low thermal conductivity and high electrical conductivity. ### Pattern Recognition thermoelectric figure of merit is given by Z = fracS^2 sigmakappa$Z = \frac{S^2 \sigma}{\kappa}$, where sigma$\sigma$ is electrical conductivity and kappa$\kappa$ is thermal conductivity. To maximize Z$Z$, we inherently need high sigma$\sigma$ and low kappa$\kappa$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity Class 11 Physics: Thermal Properties of Matter

### Core Logic Label the circuit nodes carefully to see how the elements are connected across terminal zones:

Tracing parallel loops reveals all inner subsets share identical path boundaries. mathrmR_texteq = fracmathrmr/33 = fracmathrmr9$\mathrm{R}_{\text{eq}} = \frac{\mathrm{r}/3}{3} = \frac{\mathrm{r}}{9}$ ### Step 1: Final Expression The overall simplified system network value equals fracmathrmr9$\frac{\mathrm{r}}{9}$, matching option (3). ### Pattern Recognition Short-circuit tracking lines allow collapsing complex meshes into basic parallel branches. If all nodes connect symmetrically, apply simple division: mathrmR/mathrmN$\mathrm{R}/\mathrm{N}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity