Answer:
(a) I₁ = 20/17 ≈ 1.18
I₂ = 23/17 ≈ 1.35
I₃ = 3/17 ≈ 0.18
I₄ = 1
I₅ = 14/17 ≈ 0.82
(b) 23.1 W
(c) 23.1 W
Explanation:
Kirchhoff's circuit laws have two parts: the voltage law says that the sum of all voltage drops and gains around a loop is zero, and the current law says that the sum of currents entering and leaving a node is zero.
There are 5 unknowns (I₁, I₂, I₃, I₄, I₅), so we will need 5 equations.
Looking at the bottom left node:
I₁ − I₂ + I₃ = 0
Looking at the bottom right node:
-I₃ + I₄ − I₅ = 0
Evaluating the outer loop, the sum of voltages is:
14 − 5I₄ − 5I₃ − 6I₂ = 0
Evaluating the left loop, the sum of voltages is:
14 − 5I₁ − 6I₂ = 0
Evaluating the right loop, the sum of voltages is:
5 − 5I₄ = 0
Solving the fifth equation, I₄ = 1 A.
Subtract the third equation from the fourth:
(14 − 5I₄ − 5I₃ − 6I₂) − (14 − 5I₁ − 6I₂) = 0
5I₁ − 5I₄ − 5I₃ = 0
I₁ − I₄ − I₃ = 0
I₁ = I₃ + I₄
I₁ = I₃ + 1
Substitute into the first equation:
I₃ + 1 − I₂ + I₃ = 0
1 + 2I₃ − I₂ = 0
Simplifying the third equation:
14 − 5(1) − 5I₃ − 6I₂ = 0
9 − 5I₃ − 6I₂ = 0
Eliminate I₂ by multiplying the top equation by 6 and subtracting the bottom equation.
(6 + 12I₃ − 6I₂) − (9 − 5I₃ − 6I₂) = 0
-3 + 17I₃ = 0
I₃ = 3/17
Solve for I₁:
I₁ = I₃ + 1
I₁ = 20/17
Plug into the first and second equations to find I₂ and I₅:
I₂ = I₁ + I₃ = 23/17
I₅ = -I₃ + I₄ = 14/17
(b) Power is voltage times current, so the power supplied by each voltage source is the voltage times the corresponding current.
P = ΣVI
P = (14 V) (23/17 A) + (5 V) (14/17 A)
P = 392/17 W
P ≈ 23.1 W
(c) Energy is conserved, so the power dissipated by the resistors is equal to the power supplied by the voltage sources.
P ≈ 23.1 W
