Two cells of emf 1V and 2V and internal resistance 2Omega$2\Omega$ and 1Omega$1\Omega$, respectively, are connected in series with an external resistance of 6Omega$6\Omega$. The total current in the circuit is I_1$I_{1}$ Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is I_2$I_{2}$. The value of (fracI_1I_2)$(\frac{I_{1}}{I_{2}})$ is fracx3$\frac{x}{3}$. The value of x is ________.

### Related Formula R_texteq = fracR_1 R_2R_1 + R_2 text (parallel)$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \text{ (parallel)}$$ H = I^2 R t$H = I^2 R t$ I = fracVR_texteq + r$$I = \frac{V}{R_{\text{eq}} + r}$$ ### Core Logic The circuit is a balanced Wheatstone bridge between A and B, because the ratio of adjacent arms is frac12 = frac24$\frac{1}{2} = \frac{2}{4}$. Thus, the middle 1\,Omega$1\,\Omega$ resistor (if there is one connecting the middle nodes) is ineffective.

The equivalent resistance across AB (R_AB$R_{AB}$) involves the upper branch (1+2 = 3\,Omega$1+2 = 3\,\Omega$) and the lower branch (2+4 = 6\,Omega$2+4 = 6\,\Omega$) in parallel.

R_AB = frac3 times 63 + 6 = frac189 = 2\,Omega$$R_{AB} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$$ ### Step 1: Calculate Current and Heat Total resistance in the circuit = R_AB + r_textinternal = 2 + 1 = 3\,Omega$R_{AB} + r_{\text{internal}} = 2 + 1 = 3\,\Omega$. Total current from the battery i = frac9textV3\,Omega = 3textA$i = \frac{9\text{V}}{3\,\Omega} = 3\text{A}$. The heat generated specifically between points A and B in 1 minute (60 seconds) is: H_AB = i^2 R_AB t = (3)^2 times 2 times 60$$H_{AB} = i^2 R_{AB} t = (3)^2 \times 2 \times 60$$ H_AB = 9 times 120 = 1080text J$$H_{AB} = 9 \times 120 = 1080\text{ J}$$ ### Pattern Recognition Check for balanced Wheatstone bridge first. Then use H = i^2 R t$H = i^2 R t$ strictly with the equivalent resistance of just the section AB$AB$ to find heat specifically generated across AB$AB$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### 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 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