Verified answer

V = I x R

V = .72 x 9

V = 6.48V

V = 6.5V

I'm gonna give you a general rule and idea, you should be able to do it following this rule.

The total resistance of a set of resistors in parallel is found by adding up the reciprocals of the resistance values, and then taking the reciprocal of the total:

equivalent resistance of resistors in parallel: 1 / R = 1 / R1 + 1 / R2 + 1 / R3 +...

Use the present branch technique: for 2(2) resistor that are related in parallel, say R1 and R2, and given source cutting-edge, say I and cutting-edge that previous by R1 and R2, say I1 and I2 respectively, the formula could actual be : I1 = I(R2 / (R1 + R2) ) I2 = I(R1 / (R1 + R2) ) or purely... = I - I1 So, as on your question : enable : I = 2.0 A R1 = 2.3 ohm R2 = 4.2 ohm consequently, cutting-edge pass in the two.3 ohm resistor = I1 I1 = I(R2 / (R1 + R2) ) = 2.0 (4.2 / (2.3 + 4.2) ) = a million.29 A ...

V=IR

find the total resistance (1/R)=(1/9)+(1/84)

and find the current in the second resistor (i don't remember if it splits when they're in parallel, probably says in your notes somewhere) if it does, find the second current and add them for the total to use in the formula. if it doesn't then just use .720A in the formula