The emfs of three cells connected in parallel | vedclass

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(C) For cells connected in parallel,the equivalent emf $E_{	ext{eq}}$ and equivalent internal resistance $r_{	ext{eq}}$ are given by the formula: $\

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$47$

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(C) For cells connected in parallel,the equivalent emf $E_{	ext{eq}}$ and equivalent internal resistance $r_{	ext{eq}}$ are given by the formula: $\frac{E_{	ext{eq}}}{r_{	ext{eq}}} = \sum \frac{E_i}{r_i}$ and $\frac{1}{r_{	ext{eq}}} = \sum \frac{1}{r_i}$.First,calculate the initial value of $\frac{E_{	ext{eq}}}{r_{	ext{eq}}}$:$\frac{E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{5}{1} + \frac{8}{2} + \frac{10}{3} = 5 + 4 + 3.33 = \frac{37}{3} \ 	ext{A}$.When $E_3$ is changed to $E_{3N}$,the equivalent emf becomes $2E_{	ext{eq}}$. The new equivalent internal resistance $r_{	ext{eq}}$ remains the same as it depends only on the internal resistances.Thus,the new equation is: $\frac{2E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{5}{1} + \frac{8}{2} + \frac{E_{3N}}{3} = 9 + \frac{E_{3N}}{3} = \frac{27 + E_{3N}}{3}$.Since $\frac{E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{37}{3}$,we have $\frac{2E_{	ext{eq}}}{r_{	ext{eq}}} = \frac{74}{3}$.Equating the two expressions: $\frac{27 + E_{3N}}{3} = \frac{74}{3}$.$27 + E_{3N} = 74$.$E_{3N} = 74 - 27 = 47 \ V$.

The emfs of three cells connected in parallel are $E_1=5 \ V, E_2=8 \ V$ and $E_3=10 \ V$ and their internal resistances are $r_1=1 \ \Omega, r_2=2 \ \Omega$ and $r_3=3 \ \Omega$,respectively. By changing $E_3$ to $E_{3N}$,the equivalent emf is doubled. Then $E_{3N}$ in $V$ is:

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