Question

The emfs of three cells connected in parallel are $E_1=5V,E_2=8V$ and $E_3=10V$ 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

• 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 $E_{3N}$, equivalent emf gets doubled, so new equation
$\frac{2E_{eq}}{r_{eq}}=\frac{5}{1}+\frac{8}{2}+\frac{E_{3N}}{3}$
$=\frac{27+E_{3N}}{3}$
or $\frac{E_{eq}}{r_{eq}}=\frac{27+E_{3N}}{6}$ ...(ii)
From Eqs (i) and (ii), we get
$\Rightarrow \frac{27+E_{3N}}{6}=\frac{37}{3}$
$\Rightarrow 27+E_{3N}=74$
$\Rightarrow E_{3N}=47V$