Question

In the circuit shown $E,F,G$ and $H$ are cells of e.m.f. $2V,1V,3V$ and $1V$ respectively and their internal resistances are $2\Omega ,1\Omega ,3\Omega$ and $1\Omega$ respectively.

• A. $V_D-V_B=\frac{2}{13}V$
• B. $V_D-V_B=-\frac{2}{13}V$
• C. $V_G=\frac{21}{13}V=$ potential difference across $G$
  
• D. $V_H=\frac{19}{13}V=$ potential difference across $H$

Answer 1

Answer: (B), (C), (D)

$V_H=\frac{19}{13}V=$ potential difference across $H$

Let the current through the three branches between B and D be $i_1,i_2$ and $i_3$ as shown in the figure.


Let $V_D-V_B=x$.

Then, the voltage drop across all three paths of the circuit should equal $x$.

$-3i_1+2-1=x$ -- (i)
$-2i_2=x$ -- (ii)
$4i_3+1-3=x$ -- (iii)

At point $B$,
$i_3=i_1+i_2$ (Kirchoff's current law)

Substituting from (i), (ii) and (iii),

$\frac{x+2}{4}=\frac{1-x}{3}+\frac{x}{2}$

$\Rightarrow x=-\frac{2}{13}V$

Backsubstituting the value of $x$,
$i_1=\frac{5}{13}A,i_2=\frac{1}{13}A,i_3=\frac{6}{13}A$

Potential difference across G
$V_G=3-3i_3=3-3\frac{6}{13}=\frac{21}{13}V$

$V_H=1+1\times i_3=\frac{19}{13}V$