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Series Combination of Cells

In the circui...

Question

In the circuit shown below the cells $E_{1}$ and $E_{2}$ have emf’s 4 V and 8 V and internal resistance 0.5 ohm and 1 ohm respectively. Then the potential difference across cell $E_{1}$ and $E_{2}$ will be

3.75 V, 7.5 V

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4.25 V, 7.5 V

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3.75 V, 3.5 V

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4.25 V, 4.25 V

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Solution

The correct option is B 4.25 V, 7.5 V
In the given circuit diagram external resistance $R = \frac{3 \times 6}{3 + 6} + 4.5 = 6.5 Ω$
Hence main current through the circuit $i = \frac{E_{2} - E_{1}}{R + r_{e q}} = \frac{8 - 4}{6.5 + 0.5 + 0.5} = \frac{1}{2}$ amp
Cell 1 is charging so from it’s emf equation $E_{1} = V_{1} - i r_{1} \Rightarrow 4 = V_{1} - \frac{1}{2} \times 0.5 \Rightarrow V_{1} = 4.25$ volt
Cell 2 is discharging so from it’s emf equation $E_{2} = V_{2} + i r_{2} \Rightarrow 8 = V_{2} + \frac{1}{2} \times 1 \Rightarrow V_{2} = 7.5$ volt

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In the circuit shown below the cells E1 and E2 have emf’s 4 V and 8 V and internal resistance 0.5 ohm and 1 ohm respectively. Then the potential difference across cell E1 and E2 will be

In the circuit shown below the cells $E_{1}$ and $E_{2}$ have emf’s 4 V and 8 V and internal resistance 0.5 ohm and 1 ohm respectively. Then the potential difference across cell $E_{1}$ and $E_{2}$ will be