The potential difference across A B in the ne | vedclass

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(B) In the given circuit,there are four cells each of $EMF$ $E$ and four resistors each of resistance $r$ connected in a series loop.The total $EMF$ o

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$E$

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(B) In the given circuit,there are four cells each of $EMF$ $E$ and four resistors each of resistance $r$ connected in a series loop.The total $EMF$ of the circuit is $\sum E = E - E + E - E = 0$.Since the total $EMF$ is zero,the current $i$ flowing through the circuit is $i = \frac{\sum E}{\sum R} = \frac{0}{4r} = 0$.To find the potential difference $V_{AB} = V_A - V_B$,we traverse the path from $A$ to $B$ through the resistors and cells.Starting from $A$ and moving towards $B$ along the path containing the resistor $r$ and the cell $E$ (bottom branch),we have:$V_A - i r - E = V_B$Since $i = 0$,this simplifies to $V_A - V_B = E$.Alternatively,moving along the other path (top branch) containing the cell $E$ and resistor $r$:$V_A - E + i r = V_B$$V_A - V_B = E - i r = E - 0 = E$.Thus,the potential difference across $A B$ is $E$.

The potential difference across $A B$ in the network shown is .........

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