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 - 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