Two cells of emf 1 V and 2 V and internal resistance 2 and 1 ,...

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)

\frac{x}{3}

. The value of

x

is _____.

Correct answer

Step-by-step 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}

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