Help with kirchoffs rule,......?

Question

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question number is 76

Answer 1

To start, assume a (conventional) current direction for the three branches. If you pick the wrong direction, your answers will be negative, meaning current actually flows in opposite direction.

Assumptions for the three branches.

(Top branch) E_1 = 4V with branch current I_1 leaving positive terminal (to the right). R_1 is 8Ω.

(Bottom branch) E_2 = 12V with branch current I_2 leaving positive terminal.

(Middle branch) Branch current I_3 going to the left. R_2 = 2Ω.

Do Kirchhoff's Voltage Law (KVL) around both loops. If the voltage makes a positive to negative transition or (conventional) current enters the resistor, it is negative. Positive for vice versa.

KVL Loop 1 + E_1 - V_R1 - V_R2 = 0

+ E_1 - I_1 R_1 - I_3 R_2 = 0

+4V - 8 I_1 - 2 I_3 = 0 Equation 1

KVL Loop 2 + E_2 - V_R2 = 0

+ E_2 - I_3 R_2 = 0

+ 12V - 2 I_3 = 0 Equation 2 (one unknown)

I_3 = 12V/2Ω = 6A

Substituting this into Equation 1, we get:

+4V - 8Ω I_1 - 2Ω 6A = 0

- 8Ω I_1 = 12V - 4V

I_1 = -[(12V - 4V)/8Ω] = -1A

So  I_3 = 6A and I_1 = -1A.

The negative I_1 means it is actually going to the left.  Our assumption was wrong.  

By Kirchhoff's Current Law (KCL)

+ I_1 + I_2 - I_3 = 0

I_2 =  I_3 - I_1 = 6A - (-1A) = 7A

So the bigger battery is providing current for 2Ω and attempting to charge smaller battery, which is usually bad for smaller battery.

Answer 2

http://www.physics.udel.edu/~smbarr/Phys...