Physics - Current Electricity Question with Solution | TestHub

Applying Kirchoff's loop  rule in $\mathrm{ABCDA}$

$50 – 20{\mathrm{I}}_1 = 0$

$\therefore {\text{I}}_1=2·5 \text{A}$

Applying Kirchoff's loop rule in $\mathrm{EGHFE}$

$-10{\mathrm{I}}_2+30=0$

$\therefore {\text{I}}_2=3 \text{A}$

Applying Kirchoff's loop rule in $\mathrm{CEFDC}$

$-5\left(\text{I}-{\text{I}}_1\right)-10{\text{I}}_2-5\left(\text{I}-{\text{I}}_1\right)+20{\text{I}}_1=0$

$-10\left(\text{I}-2·5\right)-30+\left(20\times 2·5\right)=0$

$-10\text{I}+25-30+50=0$

$\text{I}=4·5 \text{A}$

$\text{I}-{\text{I}}_1-{\text{I}}_2=4·5-2·5-3=-1$

$4.5 \mathrm{A}$ and $1 \mathrm{A}$