Question

1. For the circuit shown in the figure, we wish to find the currents 11, 12, and 13. Use Kirchhoffs rules to obtain equations for a. the upper loop, b. the lower loop, and c. the junction on the left side. In each case, suppress units for clarity and simplify, combining the terms. d. Solve the junction equation for Iz. e. Using the equation found in part (d), eliminate 13 from the equation found in part (b). f. Solve the equations found in parts (a) and (e) simultaneously for the two unknowns Iand 12. g. Substitute the answers found in part (f) into the junction equation found in part (d), solving for 13. h. What is the significance of the negative answer for 12? 18.0 V 5.00 22 8.002 12.0 V 11.02 7.00 22 M T, 5.00 22 13 36.0 V

Answer 1

Solution:

a. Applying Kirchhoff's Voltage Law (KVL) to the upper loop,

  –18.0+5.00I₁+7.00I₂–12.0+11.0I₂+8I₁ = 0

13I₁+18.0I₂–30=0

b. Applying Kirchhoff's Voltage Law (KVL) to the upper loop,

  –11.0I₂+12.0–7.00I₂–36.0+5.00I₃=0

  –18.0I₂++5.00I₃–24.0 = 0

c. Applying Kirchhoff's junction law to the junction on the left side,

I₁–I₂–I₃ = 0

d.    I₃ = I₁–I₂

e. Substituting the equaltion found in d in equation found in b,

   –18.0I₂++5.00( I₁–I₂)–24=0

  –18.0I₂+5.00I₁–5.00I₂–24=0

5.00I₁–23.0I₂–24=0

f. Solving the equations found in (a) and (e) for I₁ and I₂:

multiplying equation (a) with 5 and equation (e) with 13 and subtracting,

(a)x5  – (e) x 13 $\Rightarrow$ 65I₁+90.0I₂ – 150 – 65.00I₁ + 299.0I₂ + 312 = 0

   389.0I₂ + 162 = 0

I₂ = –162 / 389.0

I₂ = – 0.416 A

Substituting I₂ value in equation (a) ,

  13I₁+18.0(–0.416)–30 = 0   

13I₁ – 7.49 – 30 =0

13I₁ = 37.49

I₁ = 2.9A

Now, substituting the values of I₁ and I₂ in equation found in (d),

I₃ = 2.9 – (– 0.416)

    I₃ = 2.9+0.416

I₃ = 3.316

   I₃ = 3.3A

So, the currents in the circuit are  I₁ = 2.9A ,  I₂ = – 0.416 A, and I₃ = 3.3A.

The significance of the negative sign of the current I₂ is that the direction of I₂ is opposite to the direction we initially assumed.