Two equal resistances ${\mathrm{R}}_1={\mathrm{R}}_2=\mathrm{R}$ are connected with a $30\mathrm{\Omega }$ resistor and a battery of terminal voltage E. the currents in the two branches are 2.25 A and 1.5 A as shown in figure then

Using Kirchhoff’s 1st law current through R₂ is $2.25-1.5=0.75\mathrm{A}.$ 

Also since R₂ is in parallel with the $30\mathrm{\Omega }$ resistance, R₂ must be $60\mathrm{\Omega }$ since only half the current flows through it compared to the current through $30\mathrm{\Omega }$ resistor. 

Total resistance in the circuit becomes $60+20=80\mathrm{\Omega }$ 

Potential drop across the battery $\mathrm{E}=\mathrm{IR}=2.25\times 80=180\mathrm{V}$