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  • A cell E1 of emf 6V and internal resistance 2  is connected with another cell E2 of emf 4V and internal resistance 8<img alt="\Om

A cell E1 of emf 6V and internal resistance 2 \Omega is connected with another cell E2 of emf 4V and internal resistance 8\Omega ( as shown in the figure). The potential difference across points X and Y is:
Option: 1 3.6V
Option: 2 5.6V
Option: 3 10.0 V
Option: 4 2.0V

Answers (1)

best_answer

The figure given in the question can be redrawn as given below:-

\\ \text{The direction of current in circuit will be as shown in the figure.} \\ \text{So point}\ \mathrm{Y} \ \text{is at higher potential than}\ \mathrm{X} .\\ \mathrm{So}\ \mathrm{V}_{\mathrm{Y}}>\mathrm{V}_{\mathrm{X}} \\ \text{Current(I) in circuit,}\\ \mathbf{I}=\frac{\mathbf{E}_{1}+\mathbf{E}_{2}}{\mathbf{r}_{1}+\mathbf{r}_{2}}=\frac{(\mathbf{6}-\mathbf{4}) \mathbf{V}}{(\mathbf{2}+\mathbf{8}) \mathbf{\Omega}}=\mathbf{0 . 2} \mathbf{a m} \mathbf{p} \\\\ \text{For positive potential A is near to positive terminal of}\ \mathbf{E}_{2}\ \text{ so has}\ +\mathbf{4} \mathbf{V}

\\ \text{So potential across}\ \mathbf{E}_{1}\ and \ \mathbf{E}_{2} \\ \mathbf{E}_{1}=\mathbf{V}-\mathbf{I r}_{1}=\mathbf{6}-\mathbf{0 . 2} \times \mathbf{2}=\mathbf{6}-\mathbf{0 . 4}=\mathbf{5 . 6} \mathbf{V} \\ \mathbf{E}_{2}=\mathbf{V}+\mathbf{I r}_{2}=\mathbf{4}+\mathbf{0 . 2} \times \mathbf{8}=\mathbf{4}+\mathbf{1 . 6}=\mathbf{5 . 6} \mathbf{V} \\ \text{So potential between X and Y}=E_{2}=5.6\ Volt.

 

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