A cell E_{1} of emf 6 V and internal resistan | vedclass

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(C) The cells are connected in series. The total $emf$ of the circuit is $E_{eq} = E_{1} - E_{2} = 6 V - 4 V = 2 V$.The total resistance of the circui

Vedclass · Accepted Answer

$5.6$

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(C) The cells are connected in series. The total $emf$ of the circuit is $E_{eq} = E_{1} - E_{2} = 6 V - 4 V = 2 V$.The total resistance of the circuit is $R_{eq} = r_{1} + r_{2} = 2 \Omega + 8 \Omega = 10 \Omega$.The current $I$ flowing in the circuit is $I = \frac{E_{eq}}{R_{eq}} = \frac{2 V}{10 \Omega} = 0.2 A$.To find the potential difference across points $X$ and $Y$,we apply Kirchhoff's voltage law from $X$ to $Y$ through the cell $E_{2}$:$V_{X} - E_{2} - I \cdot r_{2} = V_{Y}$$V_{X} - V_{Y} = E_{2} + I \cdot r_{2}$$V_{X} - V_{Y} = 4 V + (0.2 A 	imes 8 \Omega) = 4 V + 1.6 V = 5.6 V$.Thus,the potential difference across points $X$ and $Y$ is $5.6 V$.

$A$ cell $E_{1}$ of $emf 6 V$ and internal resistance $2 \Omega$ is connected with another cell $E_{2}$ of $emf 4 V$ and internal resistance $8 \Omega$ as shown in the figure. The potential difference across points $X$ and $Y$ is............ $V$.

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