A circuit shown above with currents shown Write the unction

A circuit shown above with currents shown. Write the (unction rule for point A. Do not solve, simply write the equation using the notation of the graph. Write the (unction rule for point B. Do not solve, simply write the equation using the notation of the graph. Write the clockwise version of loop A starting from point A Do not solve, simplify write the equation using the notation of the graph. Write the clockwise version of loop B starting from point A do not solve simplify write the equation using the notation of the graph. Demonstrate algebraic prowess by solving for all three currents.

Solution

According to the junction law the sum ofxthe currents into the junction must equal the sum of the currents leaving the junction.

1. I₁+ I₃= I₂

2. I₂ = I₃ + I₁

According to the loop law the algebraic sum of the increases and decreases in potential difference across various components of the circuit in aclosed circuit loop must be zero.

3. I₃R₃+ 6 - I₁R₁ = 0

4. - I₃R₃ - I₂R₂+ 8 = 0

5. From equation 3

I₁ = I₃R₃+6 / R₁

From equation 4

I₃ = -I₂R₂ +8 / R₃

Sub the values I₁ and I₃ values in equation 1 we get I₂