Answer:
2.72 A
7.41 V
3.95 V
Explanation:
Let's say I₁ is the current in the 3Ω resistor, I₂ is the current in the 10Ω resistor, and I₃ is the current in the 2Ω resistor.
Apply Kirchoff's voltage law to the top loop:
20 − 6 I₂ − 10 I₂ − 3 I₁ = 0
20 − 16 I₂ − 3 I₁ = 0
Apply Kirchoff's voltage law to the bottom loop:
20 − 4 I₃ − 2 I₃ − 3 I₁ = 0
20 − 6 I₃ − 3 I₁ = 0
Apply Kirchoff's current law at either junction:
I₁ − I₂ − I₃ = 0
I₁ = I₂ + I₃
Three equations, three variables. To solve, let's multiply the first equation by 3 and the second equation by 8:
60 − 48 I₂ − 9 I₁ = 0
160 − 48 I₃ − 24 I₁ = 0
Add:
220 − 48 I₂ − 48 I₃ − 33 I₁ = 0
220 − 48 (I₂ + I₃) − 33 I₁ = 0
Substitute and solve:
220 − 48 I₁ − 33 I₁ = 0
220 − 81 I₁ = 0
I₁ = 2.72
Therefore, from the first equation:
I₂ = 0.741
And from either the second or third equation:
I₃ = 1.98
So the voltage across the 10Ω resistor is:
V = IR
V = (0.741 A) (10 Ω)
V = 7.41 V
And the voltage across the 2Ω resistor is:
V = IR
V = (1.98 A) (2 Ω)
V = 3.95 V
