Physics - Current Electricity Question with Solution | TestHub

Applying the loop law to BADB we get,

$(2\Omega ){\mathrm{i}}_1-2\mathrm{V}+1\mathrm{V}+(1\Omega ){\mathrm{i}}_1+(2\Omega )\left({\mathrm{i}}_1-{\mathrm{i}}_2\right)=0$

or $(5\Omega ){\mathrm{i}}_1-(2\Omega ){\mathrm{i}}_2=1\mathrm{V}$ ....(i)

Applying the same law to the loop DCBD, we get,

$-3\mathrm{V}+(3\Omega ){\mathrm{i}}_2+(1\Omega ){\mathrm{i}}_2+1\mathrm{V}-(2\Omega )\left({\mathrm{i}}_1-{\mathrm{i}}_2\right)=0$

or, $-(2\Omega ){\mathrm{i}}_1+(6\Omega ){\mathrm{i}}_2=2\mathrm{V}$ .....(ii)

From (i) and (ii),

$i_1=\frac{5}{13}A,i_2=\frac{6}{13}A$

So that, $i_1-i_2=-\frac{1}{13}A$

The current in BD is from B to D.

(a) ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{D}}=(2\Omega )\left(\frac{1}{13}\mathrm{A}\right)=\frac{2}{13}\mathrm{V}$

(b) The potential difference across the cell G is

${\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{D}}=-(3\Omega ){\mathrm{i}}_2+3\mathrm{V}$ $=\left(3\mathrm{V}-\frac{18}{13}\mathrm{V}\right)=\frac{21}{13}\mathrm{V}$

The potential difference across the cell H is

${\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}}=(1\Omega ){\mathrm{i}}_2+1\mathrm{V}-(1\Omega )\left(\frac{6}{13}\mathrm{A}\right)+1\mathrm{V}=\frac{19}{13}\mathrm{V}$