Determine the voltmeter reading in the given circuit.

Consider node ‘a’ as mentioned above,

Assume voltage at node ‘a’ as V_a,

Apply KCL at node ‘a’, (assuming all currents are leaving the node)

\(\frac{{{V_a}}}{{20}} + 0.25 + \frac{{\left( {{V_a} - 80{i_1}} \right)}}{{60}} = 0\)

\(\frac{{3{\rm{Va\;}} + {\rm{\;}}15{\rm{\;}} + {\rm{\;}}{\rm{Va\;}}-{\rm{\;}}80{{\rm{i}}_1}}}{{60}}{\rm{ = 0}}\)

3V_a + 15 + V_a – 80 i₁ = 0

4V_a + 15 = 80 i₁   -----(1)

From the circuit we can say,

V_a / 20 = - i₁

V_a = - 20 i₁

∴ equation(1) reduces as

4 × (- 20 i₁) + 15 = 80 i₁

\({I_{1\;}} = \frac{{15}}{{160{\rm{ }}}} A\)

Let, the voltage reads by the voltmeter = V₀

It is the voltage value of the dependent source.

V0  = 80 i₁

\({{\rm{V}}_0}{\rm{\;}} = {\rm{\;}}80{\rm{\;}} \times \frac{{15}}{{160}}\)

V0  = 7.5 V