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 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

### Related Formula R = fracVI$$R = \frac{V}{I}$$ R = R_0 (1 + alpha Delta T)$$R = R_0 (1 + \alpha \Delta T)$$ ### Core Logic First, evaluate the final resistance R_T$R_T$ at the operating condition using Ohm's law. Second, plug the final resistance into the linear temperature dependence equation for resistance to solve for final temperature T$T$. ### Step 1: Calculate Final Resistance Given V = 220 mathrm~V$V = 220 \mathrm{~V}$ and I = 2.75 mathrm~A$I = 2.75 \mathrm{~A}$: R_T = frac2202.75 = 80 \,Omega$$R_T = \frac{220}{2.75} = 80 \,\Omega$$ ### Step 2: Apply Temperature Equation We know R_27 = 60 \,Omega$R_{27} = 60 \,\Omega$. R_T = R_27 [1 + alpha (T - 27)]$$R_T = R_{27} [1 + \alpha (T - 27)]$$ 80 = 60 [1 + 2 times 10^-4 (T - 27)]$$80 = 60 [1 + 2 \times 10^{-4} (T - 27)]$$ frac8060 = 1 + 2 times 10^-4 (T - 27)$$\frac{80}{60} = 1 + 2 \times 10^{-4} (T - 27)$$ frac43 - 1 = 2 times 10^-4 (T - 27)$$\frac{4}{3} - 1 = 2 \times 10^{-4} (T - 27)$$ frac13 = 2 times 10^-4 (T - 27)$$\frac{1}{3} = 2 \times 10^{-4} (T - 27)$$ ### Step 3: Solve for T T - 27 = frac16 times 10^-4$$T - 27 = \frac{1}{6 \times 10^{-4}}$$ T - 27 = frac100006 = 1666.67$$T - 27 = \frac{10000}{6} = 1666.67$$ T = 1666.67 + 27 approx 1693.67 Rightarrow 1694^circ mathrmC$$T = 1666.67 + 27 \approx 1693.67 \Rightarrow 1694^\circ \mathrm{C}$$ ### Pattern Recognition Always separate the final temperature T$T$ from Delta T$\Delta T$. The most common error is forgetting to add back the initial reference temperature (27^circmathrmC$27^\circ\mathrm{C}$) at the end. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula I = fracE_textnetR_texteq$$I = \frac{E_{\text{net}}}{R_{\text{eq}}}$$ V = E - Ir quad (textDischarging)$$V = E - Ir \quad (\text{Discharging})$$ V = E + Ir quad (textCharging)$$V = E + Ir \quad (\text{Charging})$$ ### Core Logic

Because the cells are in opposition, the net EMF drives current from the higher potential cell (8mathrmV$8\mathrm{V}$) to the lower potential cell (2mathrmV$2\mathrm{V}$). Thus, the 2mathrmV$2\mathrm{V}$ cell acts as a load and undergoes charging. ### Step 1: Calculate Total Current I = frac8 - 22 + 4 = frac66 = 1 mathrm~A$$I = \frac{8 - 2}{2 + 4} = \frac{6}{6} = 1 \mathrm{~A}$$ ### Step 2: Terminal Potential of Cell 2 Since cell E_2$E_2$ is being charged, its terminal potential difference is: V_2 = E_2 + I r_2$$V_2 = E_2 + I r_2$$ Applying Kirchhoff's rule across cell E_2$E_2$ (from node C to B): V_C - V_B = E_2 + I r_2 = 2 + (1)(4) = 6 mathrm~V$$V_C - V_B = E_2 + I r_2 = 2 + (1)(4) = 6 \mathrm{~V}$$ ### Pattern Recognition When a smaller battery is forced backwards by a larger battery, it gets "charged". Consequently, its terminal voltage increases above its nominal EMF: V = E + Ir$V = E + Ir$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula fracR_textleftR_textwire-left = fracR_textrightR_textwire-right$$\frac{R_{\text{left}}}{R_{\text{wire-left}}} = \frac{R_{\text{right}}}{R_{\text{wire-right}}}$$ ### Core Logic For a meter bridge, the balancing condition is independent of the absolute resistance of the bridge wire as long as it is uniform. The ratio of the resistances in the gaps balances with the ratio of lengths.

### Step 1: First Condition fracXr ell_1 = frac25r (100 - ell_1)$$\frac{X}{r \ell_1} = \frac{25}{r (100 - \ell_1)}$$ Given ell_1 = 40 text cm$\ell_1 = 40 \text{ cm}$: fracXr times 40 = frac25r times 60 implies fracX40 = frac2560$$\frac{X}{r \times 40} = \frac{25}{r \times 60} \implies \frac{X}{40} = \frac{25}{60}$$ ### Step 2: Second Condition When replaced by a wire of 2r$2r$ per cm, the new lengths ell_2$\ell_2$ will satisfy: fracX2r ell_2 = frac252r (100 - ell_2)$$\frac{X}{2r \ell_2} = \frac{25}{2r (100 - \ell_2)}$$ Notice that the 2r$2r$ terms cancel out entirely from both sides, leaving: fracXell_2 = frac25100 - ell_2$$\frac{X}{\ell_2} = \frac{25}{100 - \ell_2}$$ ### Step 3: Conclusion Since the ratio X/25$X/25$ remains identical, the balancing length ratio ell / (100-ell)$\ell / (100-\ell)$ also remains identical. Therefore, ell_2 = ell_1 = 40 text cm$\ell_2 = \ell_1 = 40 \text{ cm}$. ### Pattern Recognition Meter bridge balance point strictly depends on length ratio, NOT the specific resistivity or thickness of the wire (provided it is uniform). If external resistors don't change, the balance point never changes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula Power (Illumination) is proportional to the square of the current for a constant resistance: P = I^2 R$P = I^2 R$ ### Core Logic Initial power: P_1 = I_1^2 R$P_1 = I_1^2 R$ If current drops by 20%, the new current is: I_2 = I_1 - 0.2 I_1 = 0.8 I_1$$I_2 = I_1 - 0.2 I_1 = 0.8 I_1$$ New power: P_2 = (0.8 I_1)^2 R = 0.64 I_1^2 R = 0.64 P_1$$P_2 = (0.8 I_1)^2 R = 0.64 I_1^2 R = 0.64 P_1$$ ### Step 1: Calculate Percentage Change Delta P \% = fracP_2 - P_1P_1 times 100\%$$\Delta P \% = \frac{P_2 - P_1}{P_1} \times 100\%$$ Delta P \% = frac0.64 P_1 - P_1P_1 times 100\%$$\Delta P \% = \frac{0.64 P_1 - P_1}{P_1} \times 100\%$$ Delta P \% = (0.64 - 1) times 100\% = -36\%$$\Delta P \% = (0.64 - 1) \times 100\% = -36\%$$ ### Step 2: Final Statement The negative sign indicates a decrease. The illumination drops by 36\%$36\%$. ### Pattern Recognition For squared relations y = x^2$y = x^2$, if x$x$ changes by a factor k$k$ (0.8$0.8$), y$y$ changes by a factor k^2$k^2$ (0.64$0.64$). 1 - 0.64 = 36\%$1 - 0.64 = 36\%$. This bypasses algebraic limits usually done for small changes (like 2Delta x / x$2\Delta x / x$) since 20% is too large for approximation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity