Charge that flows in a circuit encounters a change in electric potential (or electric pressure) whenever it passes through a resistor such as a bulb or a battery. There is a gain in electric potential within the battery and a loss of electric potential within the resistors (e.g., the bulbs). The loss in electric potential - sometimes referred to as a voltage drop (∆V) - can be calculated from knowledge of the current (I) and the resistance (R). The product I•R is equal to the voltage drop. Knowing the battery voltage and the current and resistance for the bulbs, one can calculate the voltage drops and thus the electric potential for each location in a circuit.
 

The battery has a voltage rating of 36 V. This is the difference in electric potential between the positive and the negative terminal. Since the negative terminal is typically assigned a potential of 0 V, the positive terminal of this battery is 36 V higher in potential. Any wire attached to these terminals shares the same electric potential value as the terminal it is attached to. This allows you to determine the electric potential at locations A and D.  

Charge encounters a voltage drop (or loss of electric potential) as it passes through a bulb. The voltage drop is calculated as I•R. The I is listed in amps (A) and the R is listed in ohms (Ω),  Multiplying these two values provides knowlege of the amount of voltage drop (∆V). Subtracting this ∆V value from the electric potential value just prior to the bulb allows one to determine the electric potential value just after the bulb. This mulitplying and subtracting method will have to be done for both bulbs to determine the electric potential at locations B and C.

You can easily do a final check of your calculations. A third set of I an R values are given for the third bulb. Thus, you can calculate a third ∆V value. This ∆V should be equal to the difference in electric potential values between locations C and D.
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