RESISTORS

1) Using the kirchoff's first law, we have I1 = I2 + I3

Then applying Kirchoff's second law for (with all r = 1 ohm)

The upper loop: 24 = I1 + 22I1 + 12I1 + I2 + 11I2

24 = 35I1 + 12I2 = 25(I2 + I3) + 12I2 = 37I2 + 25I3 ... (1)

The lower loop: 6 = I2 + 11I2 - 15I3 - I3 - 18I3

6 = 12I2 - 34I3

3 = 6I2 - 17I3 ... (2)

Solving we have I3 = 0.042 A

Hence terminal voltage of the 6-V battery = 6 - 0.042 x 1 = 5.96 V

2) For the 2.6 kΩ resistor, max. tolerable current is √(0.4/2600) = 0.012 A

For the 3.6 kΩ resistor, max. tolerable current is √(0.4/3600) = 0.011 A

For the 1.8 kΩ resistor, max. tolerable current is √(0.4/1800) = 0.015 A

Hence in the connection stated in the question, we should limit the current to the consideration of the 1.8 kΩ resistor, i.e. current flow in the main branch = 0.015 A

Then suppose that current throuth the 2.6 kΩ resistor is I, we have current through the 3.6 kΩ resistor is 0.015 - I, then:

2600I = 3600(0.015 - I)

I = 0.0087 A

with the common voltage across the 2.6 kΩ and 3.6 kΩ resistors = 2600 x 0.087 = 22.5 V

In addition to the voltage across the 1.8 kΩ resistor which is 0.015 x 1800 = 26.8 V, total voltage = 22.5 + 26.8 = 49.3 or 49 V (correct to the nearest integer)

Try to use Kirchoff's 2nd law (Loop rule)