Question

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 $E_{3 N}$, the equivalent emf is doubled, then $E_{3 N}$ in $V$ is

• A. 12
• B. 34
• C. 47
• D. 82

Answer 1

Answer: (C)

Equivalent emf and resistance in parallel combination is given by
$\frac{E_{ eq }}{r_{ eq }}=\frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}}+\frac{E_{3}}{r_{3}}$
$\frac{E_{ eq }}{r_{ eq }}=\frac{5}{1}+\frac{8}{2}+\frac{10}{3}=\frac{37}{3}$ volt ...(i)
When $E_{3}$ is replaced by $E_{3 N}$, equivalent emf gets doubled, so new equation
$\frac{2 E_{ eq }}{r_{ eq }}=\frac{5}{1}+\frac{8}{2}+\frac{E_{3 N}}{3}$
$=\frac{27+E_{3 N}}{3}$
or $\frac{E_{ eq }}{r_{ eq }}=\frac{27+E_{3 N}}{6}$ ...(ii)
From Eqs (i) and (ii), we get
$\Rightarrow \frac{27+E_{3 N}}{6}=\frac{37}{3}$
$\Rightarrow 27+E_{3 N}=74$
$\Rightarrow E_{3 N}=47\, V$