JEE MAIN - Physics (2025 - 3rd April Evening Shift - No. 23)
Two cells of emfs 1 V and 2 V and internal resistances $2 \Omega$ and $1 \Omega$, respectively, are connected in series with an external resistance of $6 \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $\mathrm{I}_2$. The value of $\left(\frac{\mathrm{I}_1}{\mathrm{I}_2}\right)$ is $\frac{x}{3}$. The value of $x$ is___________.
Answer
4
Explanation
_3rd_April_Evening_Shift_en_23_1.png)
$$\begin{aligned} & \varepsilon_{\mathrm{cq}}=3 \\ & \mathrm{R}_{\mathrm{cq}}=9 \\ & \mathrm{i}_1=\frac{3}{9}=\frac{1}{3} \end{aligned}$$
_3rd_April_Evening_Shift_en_23_2.png)
$$\begin{aligned} & \varepsilon_{\text {eq }}=\frac{\frac{\varepsilon_1}{\mathrm{r}_1}+\frac{\varepsilon_2}{\mathrm{r}_2}}{\frac{1}{\mathrm{r}_1}+\frac{1}{\mathrm{r}_2}} \\ & \varepsilon_{\mathrm{eq}}=\frac{\frac{1}{2}+\frac{2}{1}}{\frac{1}{2}+\frac{1}{1}}=\frac{5}{3} \\ & \mathrm{r}_{\mathrm{equ}}=\frac{2 \times 1}{3}+6=\frac{20}{3} \\ & \mathrm{i}_2=\frac{1}{4} \Rightarrow \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{4}{3} \end{aligned}$$
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