The given circuit can be simplified to the final form shown in Fig. (d). As seen, current supplied by the battery is 1 A. At point A in Fig. (b), this current is divided into two equal parts of 0.5 A each.
Obviously, voltage V represents the potential of point B with respect to the negative terminal of the battery. Point B is above the ground by an amount equal to the voltage drop across the series combination of (40 + 50) = 90Ω.
