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 ________.

### 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

### Related Formula mathrmR = fracrhoellmathrmA implies mathrmR propto ell$$\mathrm{R} = \frac{\rho\ell}{\mathrm{A}} \implies \mathrm{R} \propto \ell$$ ### Core Logic For the equilateral triangle, the total resistance mathrmR$\mathrm{R}$ is divided into three equal line parts of value fracmathrmR3$\frac{\mathrm{R}}{3}$ each:

An edge configuration places one part in parallel with the other two series steps: left(mathrmR_texteqright)_1 = frac(frac2mathrmR3) cdot (fracmathrmR3)frac2mathrmR3 + fracmathrmR3 = frac29mathrmR$$\left(\mathrm{R}_{\text{eq}}\right)_1 = \frac{(\frac{2\mathrm{R}}{3}) \cdot (\frac{\mathrm{R}}{3})}{\frac{2\mathrm{R}}{3} + \frac{\mathrm{R}}{3}} = \frac{2}{9}\mathrm{R}$$ For the square, the wire resistance mathrmR$\mathrm{R}$ divides into four parts of fracmathrmR4$\frac{\mathrm{R}}{4}$ each:

An edge calculation pairs one section in parallel with the remaining three loops in series: left(mathrmR_texteqright)_2 = frac(frac3mathrmR4) cdot (fracmathrmR4)frac3mathrmR4 + fracmathrmR4 = frac316mathrmR$$\left(\mathrm{R}_{\text{eq}}\right)_2 = \frac{(\frac{3\mathrm{R}}{4}) \cdot (\frac{\mathrm{R}}{4})}{\frac{3\mathrm{R}}{4} + \frac{\mathrm{R}}{4}} = \frac{3}{16}\mathrm{R}$$ Taking the ratio between these geometric layouts: ### Step 1: Final Ratio Calculation fracleft(mathrmR_texteqright)_1left(mathrmR_texteqright)_2 = fracfrac29mathrmRfrac316mathrmR = frac3227$$\frac{\left(\mathrm{R}_{\text{eq}}\right)_1}{\left(\mathrm{R}_{\text{eq}}\right)_2} = \frac{\frac{2}{9}\mathrm{R}}{\frac{3}{16}\mathrm{R}} = \frac{32}{27}$$ This selects option (4). ### Pattern Recognition For a regular polygon with mathrmN$\mathrm{N}$ identical sides formed from a wire of resistance mathrmR$\mathrm{R}$, the resistance across one edge is always given by frac(mathrmN-1)mathrmN^2mathrmR$\frac{(\mathrm{N}-1)}{\mathrm{N}^2}\mathrm{R}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula Resistance of a uniform wire: R = fracrho LA$$R = \frac{\rho L}{A}$$ Equivalent resistance of two identical resistors in parallel: R_texteq = fracR_1 R_2R_1 + R_2 = fracR_texthalf2$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{R_{\text{half}}}{2}$$ ### Core Logic First, find the total resistance R_texttotal$R_{\text{total}}$ of the straight wire: - Length of wire, L = 25mathrm~m$L = 25\mathrm{~m}$ - Area of cross-section, A = 5mathrm~mm^2 = 5 times 10^-6mathrm~m^2$A = 5\mathrm{~mm}^2 = 5 \times 10^{-6}\mathrm{~m}^2$ - Resistivity, rho = 2 times 10^-6Omegamathrm~m$\rho = 2 \times 10^{-6}\Omega\mathrm{~m}$ R_texttotal = fracrho LA = frac2 times 10^-6 times 255 times 10^-6 = 10Omega$$R_{\text{total}} = \frac{\rho L}{A} = \frac{2 \times 10^{-6} \times 25}{5 \times 10^{-6}} = 10\Omega$$ ### Step 1: Splitting into Semicircles When the wire is bent into a complete circle, measuring the resistance between two diametrically opposite points splits the wire into two parallel halves of equal length. Each semicircle has a resistance of: R_textsemi = fracR_texttotal2 = frac102 = 5Omega$$R_{\text{semi}} = \frac{R_{\text{total}}}{2} = \frac{10}{2} = 5\Omega$$ These two halves are connected in parallel between the diametric terminals: R_texteq = fracR_textsemi2 = frac52 = 2.5Omega$$R_{\text{eq}} = \frac{R_{\text{semi}}}{2} = \frac{5}{2} = 2.5\Omega$$

### Step 2: Conclusion & Discrepancy The actual mathematically rigorous answer is 2.5Omega$2.5\Omega$. Since 2.5Omega$2.5\Omega$ is not present in the given options, the question is marked as a **Bonus** question by standard key evaluation guidelines. If forced to choose a theoretical option due to printing mistakes, some keys may relate it to 10Omega / 4 = 2.5Omega$10\Omega / 4 = 2.5\Omega$, but scientifically it stands as a bonus. ### Pattern Recognition Shortcut: A wire of total resistance R$R$ bent into a circle has an effective resistance across its diameter equal to R_texteq = R/4$R_{\text{eq}} = R/4$. Memorize this ratio! Here R = 10Omega$R = 10\Omega$, so R_texteq = 10/4 = 2.5Omega$R_{\text{eq}} = 10/4 = 2.5\Omega$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula Parallel combination of ammeter (R_G = 240Omega$R_G = 240\Omega$) and shunt (S = 10Omega$S = 10\Omega$): R_textparallel = fracR_G cdot SR_G + S$$R_{\text{parallel}} = \frac{R_G \cdot S}{R_G + S}$$ Total equivalent resistance: R_texteq = R_textseries + R_textparallel$$R_{\text{eq}} = R_{\text{series}} + R_{\text{parallel}}$$ Current Divider Rule: I_textammeter = I_texttotal cdot left(fracSR_G + Sright)$$I_{\text{ammeter}} = I_{\text{total}} \cdot \left(\frac{S}{R_G + S}\right)$$ ### Core Logic Let's first calculate the equivalent resistance of the parallel branch (ammeter + shunt): R_textparallel = frac240 times 10240 + 10 = frac2400250 = 9.6Omega$$R_{\text{parallel}} = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6\Omega$$ Now, add the series resistance (150.4Omega$150.4\Omega$) to find the total equivalent resistance of the circuit: R_texteq = 150.4 + 9.6 = 160Omega$$R_{\text{eq}} = 150.4 + 9.6 = 160\Omega$$ ### Step 1: Finding Total Current Use Ohm's Law to calculate the total current I_texttotal$I_{\text{total}}$ drawn from the 20mathrm~V$20\mathrm{~V}$ battery: I_texttotal = fracVR_texteq = frac20160 = 0.125mathrm~A$$I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{20}{160} = 0.125\mathrm{~A}$$ ### Step 2: Calculating Ammeter Current Using the Current Divider Rule, calculate the current I_textammeter$I_{\text{ammeter}}$ flowing through the 240Omega$240\Omega$ ammeter branch: I_textammeter = I_texttotal times frac10240 + 10 = 0.125 times frac10250$$I_{\text{ammeter}} = I_{\text{total}} \times \frac{10}{240 + 10} = 0.125 \times \frac{10}{250}$$ I_textammeter = 0.125 times frac125 = 0.005mathrm~A$$I_{\text{ammeter}} = 0.125 \times \frac{1}{25} = 0.005\mathrm{~A}$$ Convert this into milliamperes (mathrmmA$\mathrm{mA}$): I_textammeter = 0.005 times 1000 = 5mathrm~mA$$I_{\text{ammeter}} = 0.005 \times 1000 = 5\mathrm{~mA}$$ Therefore, the reading of the ammeter is 5mathrm~mA$5\mathrm{~mA}$.

### Pattern Recognition Notice how nicely the numbers are set up in JEE Mains questions! The parallel combination of 240Omega$240\Omega$ and 10Omega$10\Omega$ yields exactly 9.6Omega$9.6\Omega$, which beautifully combines with 150.4Omega$150.4\Omega$ to form a perfect integer sum of 160Omega$160\Omega$. This tells you that your intermediate steps are absolutely correct! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity: Measuring Devices

### Related Formula P = fracV^2R$$P = \frac{V^2}{R}$$ For series combination: R_s = R_1 + R_2 + dots + R_n$$R_s = R_1 + R_2 + \dots + R_n$$ ### Core Logic Since the bulbs are identical, each has a resistance R = fracV^2p$R = \frac{V^2}{p}$. When n$n$ identical bulbs are connected in series, the total equivalent resistance becomes: R_s = nR$R_s = nR$ ### Step 1: Calculate Combined Power The total power drawn across the same mains supply V$V$ is: P_s = fracV^2R_s = fracV^2nR = frac1n left(fracV^2Rright) = fracpn$$P_s = \frac{V^2}{R_s} = \frac{V^2}{nR} = \frac{1}{n} \left(\frac{V^2}{R}\right) = \frac{p}{n}$$ ### Pattern Recognition Identical appliances connected in series scale down their combined power inversely with count (P_net = P/n$P_{net} = P/n$), identical appliances in parallel scale up combined power linearly (P_net = nP$P_{net} = nP$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity