Answer
The current is the same at each point, but the potential differences are different:
$I_A = 2$ A; $V_A = 20$ Volts
$I_B = 2$ A; $V_B= 16$ Volts
$I_C = 2$ A; $V_C= 10$ Volts
$I_D = 2$ A; $V_D= 8$ Volts
$I_E = 2$ A; $V_E= 0$ Volts
Work Step by Step
Since the resistors are in series, we can find the total resistance by:
$R_{total} = \Sigma_i^n R_i= R_1+R_2+...+R_n$
$R_{total} = 2+5+4+1 = 10$ $\Omega$
and thus,
$\Delta V_{total} = IR = 2(10) = 20$ Volts
Since the current into each resistor must be the same as the amount of current flowing out of the resistor (Kirchoff's junction law) as well as having no breaks in the wire (again, the wire is in series), the current at each point is the same.
We can find the potential difference at each point by subtracting the potential difference of each individual resistor from the total potential difference.
$\Delta V_{AB} = -IR = -2(2) = -4$ Volts
$\Delta V_{BC} = -IR = -2(3) = -6$ Volts
$\Delta V_{CD} = -IR = -2(1) = -2$ Volts
$\Delta V_{DE} = -IR = -2(4) = 8$ Volts
Now we can find the potential difference at each point
$V_A = V_{total} = 20 $ Volts
$V_B = V_{total} + V_{AB} = 20 - 4 = 16 $ Volts
$V_C = V_{total} + V_{AB} + V_{BC} = 20 - 4 -6 = 10 $ Volts
$V_D = V_{total} + V_{AB} +V_{BC}+V_{CD}= 20 - 4 -6-2= 8 $ Volts
$V_D = V_{total} + V_{AB} +V_{BC}+V_{CD}-V_{DE}= 20 - 4 -6-2-8= 0 $ Volts
