Question

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

• A. $V_D-V_B=-2/13V$
• B. $V_D-V_B=2/13V$
• C. $V_G=21/13V$
• D. $V_H=19/13V$
  

Answer 1

The correct option is B $V_D-V_B=2/13V$

Given: a circuit where E,F,G and H are cells of e.m.f. $2V,1V,3V$ and $1V$ respectively and their resistances are $2\Omega ,1\Omega ,3\Omega$ and $1\Omega$ respectively.

Solution:

The current distribution in the circuit is as shown in above figure

Applying Kirchhoff’s first law at point D, we have

$i=i_1+i_2⟹i-i_1=i_2....\left(i\right)$

Applying Kirchhoff’s second law to mesh and ABDA, we have

$2i+i+2i_1=2-1⟹3i+2i_1=\mathrm{1....}\left(ii\right)$

Applying Kirchhoff’s second law to mesh DCBD, we get

$3i_2-1i_2-2i_1=3-1⟹4i_2-2i_1=\mathrm{2...}\left(iii\right)$

multiply eqn(i) by 3 and subtract eqn(ii) from it, we get

$3i-3i_1=3i_2$

$\underset{–}{-(3i+2i_1=1)}-5i_1=3i_2-1⟹3i_2+5i_1=\mathrm{1...}\left(iv\right)$

Multiply eqn(iii) by 3 and eqn(iv) by 4 and subtract them, we get

$12i_2-6i_2=6$

$\underset{–}{-(12i_2+20i_1=4)}-26i_1=-2⟹i_1=\frac{1}{13}A$

substituting the value of $i_1$ in eqn(iii), we get

$4i_2-2\left(\frac{1}{13}\right)=2⟹4i_2=2+\frac{2}{13}⟹4i_2=\frac{28}{13}⟹i_2=\frac{7}{13}A$

Potential difference between B and D

$=2i_1=\frac{2}{13}V$