In the network shown below, points A, B and C are at the potential of 70 V, 0 V and 10 V respectively. The potential of point D is
40 V
Applying KCL at junction D, VA−VD10=VD−VB20+VD−VC30⇒70−VD10=VD−020+VD−1030⇒70−VD=VD2+VD3−103⇒11VD6=2203∴VD=40 V