◆dc circuits ◆

someone with the above two questions !

12 is an absolute devil. The short answer I get is R = 1 ohm.
Step One
Draw the current loops. Call i1 the current that goes through the right hand loop which consists of 2 and 4 ohm resistors and the 6 and 10 volt batteries. Draw the loop for i1 in a clockwise direction. 

i2 is the second loop. It also goes clockwise. It goes through R 4 volts 6 volts and the 4 ohm resistor.

Step Two
Construct the current equation for loop one. 
2i1 + 4*(i1 - i2) +6 - 10 = 0
2i1 + 4*i1 - 4*i2 - 4 = 0 
6*i1 - 4*i2 - 4 = 0
3*i1 - 2*i2 - 2 = 0

Step three
Construct the i2 loop.
4(i2 - i1) + i2*R + 4 - 6 = 0
4*i2 - 4*i1 + i2*R - 2 = 0
(4 + R)i2 - 4i1 - 2 = 0

Note: the only way this is going to work is if i2 = i1. The reason for that is that that is what is going through the 4 ohm resistor. 

If that is true then going back to the first loop, we find that 
3i1 - 2*i1 - 2 = 0
i1 = 2 amps.

Step three
Find R. Use the second equation
i2 = i1 
(4 + R)*2 - 4*2  - 2  = 0
(4 + R)*2 - 8 - 2 = 0
(4 + R)*2 = 10     Divide by 2
4 + R = 5
R = 1              <<<<< Answer.

dd

It's have to be 12 or idk