Need help calculating current and potential difference

For the circuit shown in the figure (ε = 6.50 V, R = 8.00 ), calculate the following quantities.

http://www.webassign.net/pse/p28-27alt.g…



(a) Calculate the current in the 2.00 resistor.

(b) Calculate the potential difference between points a and b.



extremely frustrated nothing like this was discussed in class. please help best answer gets 10pts immediately

For simplicity we can always choose one node and assign it 0V (gnd). Let's ground the "b" node. Then we can write Kirchoff zero-sum current equation at the "a" node:

(a-12)/4 + a/2 + (a+6.5)/8 = 0

Solving: a = 2.5 Volts

Since we took b at zero volts, the potential difference between a and b is 2.5 Volts as well.

The current in the 2-Ohm resistor is 2.5/2 = 1.25 Amps

let's assume current though R resister =i1 (Assume current flowing point a to b)
let's assume current though 2 resister =i2 (Assume current flowing point b to a)
therefore current though 4ohm resister =i1 - i2 (So (i1-i2) current flow point b to a)
put Kirchhoff law for upper part circuit
12= 2(-i2)+4(i1-i2)

12=4(i1)-6(i2)----->(1)
put Kirchhoff law for lower part circuit
6.5=8(i1) +2(i2)------(2)
solving these equation
i1=(9/8)
i2=(-5/4)
that's mean actually current flowing from 2 ohm resister point a to b

so answer for question a 1.25A
potential difference between points a and b = voltage drop between 2 ohm resister=Va-Vb=2*1.25=2.5V

so answer for question b 2.5V (Only ask difference.Not ask Calculate the potential between points a and b )
always current flow from high potential to low potential (high voltage to low voltage)
therefore here Va>Vb current flowing though

u can calculate it using below circuit part too
lets start from point a negative to positive voltage increase by 6.5V
then current i1 flow point a to b so there is voltage drop 8*(9/8)=9V
therefore actually there is a voltage drop -6.5+9=2.5V (a to b)
now u can try to calculate it using upper circuit part too