Question

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

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

Answer 1

Answer: (A), (C), (D)

$V_D-V_B=-2/13V$
$V_G=21/13V=$ potential difference across $G$
$V_H=19/13V=$ potential difference across $H$

Applying Kirchhoff's law

for loop AFDBEA: $E_F-E_E=i_1(r_F+r_E+R)-R{i}_2$

or $1-2=i_1(1+2+2)-2i_2\Rightarrow 5i_1-2i_2=-1...\left(1\right)$

for loop CGDBHC: $E_H-E_G=i_2(r_H+r_G+R)-R{i}_1$

or $1-3=i_2(1+3+2)-2i_1\Rightarrow -2i_1+6i_2=-2....\left(2\right)$

$\left(1\right)\times 3\Rightarrow 15i_1-6i_2=-3...\left(3\right)$

$\left(2\right)+\left(3\right),13i_1=-5\Rightarrow i_1=-\frac{5}{13}A$

putting the value of $I_1$ in (1), we get $I_2=-\frac{6}{13}$

$V_D-V_B=$ potential drop across $R=2\Omega$ is $R(i_2-i_1)=2[-\frac{6}{13}+\frac{5}{13}]=-\frac{2}{13}V$

Thus, $V_G=E_G+i_2{r}_G=3+3(-\frac{6}{13})=\frac{21}{13}V$ (as current

drawn from cell and $i_2$ are opposite direction)

and $V_H=E_H-i_2{r}_H=1-(-\frac{6}{13})1=1+\frac{6}{13}=\frac{19}{13}V$ (as

current drawn from cell and $i_2$ are same direction.