For the network determine the voltage

for the network of fig 5.98 determine the voltges: Va, Vb, Vc,Vd, Ve.(here is the link for the circuit http://apachepersonal.miun.se/~bornor/ee/Uppg5.pdf ) (its number 27)
I know that Va is 20V and Ve is 0V, but i cant find Vb,Vc,Vd, can some one help me please.
Thank you

total battery voltage (taking CW as positive) is 20–47 = –27 volts
total R is 9k
total current is –27/9 = –3 mA, ie, flows CCW

Voltage across 2k is 6 volts, + on the right
voltage across 3k is 9 volts, + on the right
voltage across 4k is 12 volts, + down
and they add up to 27, as a check.

going CW from left ground point
Va = 20 volts
Vb = +20 + 6 = 26 volts
Vc = 26 + 9 = 35 volts
Vd = 35 – 47 = – 12 volts
and Vd checks as the voltage across the 4k

the fig. is 5.96.

current flowing through 2ohms and 3ohms resistors will be same (since both are in series). So let's call it I.


also,

Va= 20V

Vc= 47V

Ve= 0V


I = voltage/ resistance= (20-47) / (2+3) = - 5.4 A .

current is in -ve because WE HAVE ASSUMED ITS DIRECTION FROM POINT A TO C, but it will flow from higher to lower voltage (47V to 20V). so the actual flow of current is from c to a.

=> Vb= 20 - (-5.4*2) = 30.8V.

I didn't get Vd.

It shouldn't be 0 V. maybe Vd is -47V.