SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Consider node 'a' as mentioned above,
Assume voltage at node 'a' as ${\mathrm{V}}_{\mathrm{a}}$,
Apply $\mathrm{KCL}$ at node 'a', (assuming all currents are leaving the node)
$\frac{{\mathrm{V}}_{\mathrm{a}}}{20}+0.25+\frac{\left({\mathrm{V}}_{\mathrm{a}}-80{\mathrm{i}}_1\right)}{60}=0$
$\frac{3{\mathrm{V}}_{\mathrm{a}}+15+{\mathrm{V}}_{\mathrm{a}}-80{\mathit{i}}_1}{60}=0$
$3{\mathrm{V}}_{\mathrm{a}}+15+{\mathrm{V}}_{\mathrm{a}}-80{\mathit{i}}_1=0$
$4{\mathrm{V}}_{\mathrm{a}}+15=80{\mathit{i}}_1$ ......(1)
From the circuit we can say,
$\frac{{\mathrm{V}}_{\mathrm{a}}}{20}=-{\mathit{i}}_1$
${\mathrm{V}}_{\mathrm{a}}=-20{\mathit{i}}_1$
∴ equation(1) reduces as
$4\times \left(-20{\mathit{i}}_1\right)+15=80{\mathit{i}}_1$
${\mathrm{I}}_1=\frac{15}{160}\mathrm{A}$
Let, the voltage reads by the voltmeter $={\mathrm{V}}_0$
It is the voltage value of the dependent source.
${\mathrm{V}}_0=80{\mathit{i}}_1$
${\mathrm{V}}_0=80\times \frac{15}{160}$
${\mathrm{V}}_0=7.5\mathrm{V}$