Question

◆ DC Circuits ◆Someone please help with the above two questions !

Answer 1

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

Answer 2

Here would be a good place to make use of the Kirchoff Loop Rule, which states that the integral of the voltage around a closed loop of the electric field is zero (assuming no induced voltage). Simply put, the sum of voltage around any loop is zero.
Also take into account ohm's law, V = IR.

Since there is no current in the 4 ohm resistor, then there is no current throughout the entire middle strip. There is only one current around the whole circuit. Thus, Kirchoff's Junction Rule would not be very useful here.

Let's analyze the left-half loop clockwise: 10V - 2ohm*I - 6V = 0, or 4V = 2ohm*I. We can deduce that I, the current is equal to 2 A.

Now that we know the current, we can plug this in to the right-half loop, again clockwise: 6V - 2A * R - 4V = 0, or 2V = 2A*R. Now we can see that R must equal 1 ohm.

13. We can see that the 2 ohm resistor closest to the voltage source is in series with the combination of the rest of the resistors. The current through this first 2 ohm resistor is 10V/2ohm = 5 A. 

The two 3 ohm resistors are in a parallel arrangement with the second 2 ohm resistor. Combined, this arrangement also has 5 A. Inside this arrangement, the 3 ohm resistors are in series with each other, so they share the same current. The equivalent resistance of the 3 ohm resistors in series is 3 + 3 = 6 ohms.

The equivalent resistance of the 3 rightmost resistors is then 1/(1/6 + 1/2) = 1.5 ohms. 

The rightmost 3 resistors in parallel will receive a voltage equal to 10V * (1.5 ohms) / (1.5 + 2 ohms) = 30/7 V.

Within the parallel arrangement, each branch receives an equal 30/7 volts. Both 3 ohm resistors share this voltage, and since they are of equal resistance, each receives 15/7 volts. Using ohm's law, (15/7V / 3 ohms)  = 5/7 A of current passing through each 3 ohm resistor.