Two cells of emf 1 V and 2 V and internal resistance $2 \Omega$ and $1 \Omega$, respectively, are connected…

Two cells of emf 1 V and 2 V and internal resistance $2 \Omega$ and $1 \Omega$, respectively, are connected in series with an external resistance of $6 \Omega$. The total current in the circuit is $\mathrm{I}_1$. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $I_2$. The value of $\left(\frac{I_1}{I_2}\right)$ is $\frac{x}{3}$. The value of $x$ is _____.

Solution


$\begin{aligned} & \varepsilon_{\mathrm{eq}}=3 \\ & \mathrm{R}_{\mathrm{eq}}=9 \\ & \mathrm{i}_1=\frac{3}{9}=\frac{1}{3}\end{aligned}$
$\varepsilon_{\mathrm{eq}}=\frac{\frac{\varepsilon_1}{\mathrm{r}_1}+\frac{\varepsilon_2}{\mathrm{r}_2}}{\frac{1}{\mathrm{r}_1}+\frac{1}{\mathrm{r}_2}}$
$\varepsilon_{\mathrm{eq}}=\frac{\frac{1}{2}+\frac{2}{1}}{\frac{1}{2}+\frac{1}{1}}=\frac{5}{3}$
$\begin{aligned} & \mathrm{r}_{\text {equ }}=\frac{2 \times 1}{3}+6=\frac{20}{3} \\ & \mathrm{i}_2=\frac{1}{4} \Rightarrow \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{4}{3}\end{aligned}$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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