Relative to the battery, the 820 and 680 resistors combine in parallel to form a smaller resistor so

(1/820) + (1/680) = (1/R1)

1/371.7333 ohms = 1/R1

R1 = 371.7 ohms

R1 and R2 are in series so

Rt = 371.7 + 960 = 1331.7 ohms = equivalent or total resistance

To find voltage across the resistors, you're first going to need the current in the circuit.

V = IR

I = Vt/Rt = 12 V/1331.7 ohms = 9.0110E-3 A

You can determine voltage using V = IR

so start with R1 (note that since the two resistors to the left are in parallel, they will share voltage V1 which = IR1)

V1 = (9.0110E-3 A) (371.7 ohms) = 3.3493 volts

from here you can say voltage on the 960 resistor is 12 - 3.3493 = 8.6507 volts but I'm going to calculate just to be safe

V2 = (9.0110E-3 A) (960 ohms) = 8.65056--close enough, lol

a) 1331.7 ohms

b) 3.3493 volts

c) 8.6507 volts just to stay consistent with the R1

d) 3.3493 volts

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