For the network shown in the following figure the value of V_s which makes I₀ = 7.5 mA is:

Calculation:

The circuit can be redrawn as shown in the figure by taking two 6Ω in parallel whose

equivalent will be 3 Ω 

Again two  7Ω  in parallel will give 3.5Ω 

Current i = Vs/23.5 A

Now since given I₀  current flow across the last 6Ω resistor, the same current will flow across 6Ω  

connected in parallel with it. Net current flowing across 4Ω  will be 2I0

The overall current supplied by Vs voltage source will be 4 which will be \(4\times7.5=30 mA\)

\(30\times10^{-3}=Vs/23.5 \)

⇒\(V_S=705 mV\)