Yes yes ! How right you are ! Truer words were never spoken.
Answer:
Bi. Current in 15.4 Ω (R₁) is 7.14 A.
Bii. Current in 21.9 Ω (R₂) is 5.02 A.
Biii. Current in 11.7 Ω (R₃) is 9.40 A.
C. Total current in the circuit is 21.56 A.
Explanation:
Bi. Determination of the current in 15.4 Ω (R₁)
Voltage (V) = 110 V
Resistance (R₁) = 15.4 Ω
Current (I₁) =?
V = I₁R₁
110 = I₁ × 15.4
Divide both side by 15.4
I₁ = 110 / 15.4
I₁ = 7.14 A
Therefore, the current in 15.4 Ω (R₁) is 7.14 A.
Bii. Determination of the current in 21.9 Ω (R₂)
Voltage (V) = 110 V
Resistance (R₂) = 21.9 Ω
Current (I₂) =?
V = I₂R₂
110 = I₂ × 21.9
Divide both side by 21.9
I₂ = 110 / 21.9
I₂ = 5.02 A
Therefore, the current in 21.9 Ω (R₂) is 5.02 A
Biii. Determination of the current in 11.7 Ω (R₃)
Voltage (V) = 110 V
Resistance (R₃) = 11.7 Ω
Current (I₃) =?
V = I₃R₃
110 = I₃ × 11.7
Divide both side by 11.7
I₃ = 110 / 11.7
I₃ = 9.40 A
Therefore, the current in 11.7 Ω (R₃) is 9.40 A.
C. Determination of the total current.
Current 1 (I₁) = 7.14 A
Current 2 (I₂) = 5.02 A
Current 3 (I₃) = 9.40 A
Total current (Iₜ) =?
Iₜ = I₁ + I₂ + I₃
Iₜ = 7.14 + 5.02 + 9.40
Iₜ = 21.56 A
Therefore, the total current in the circuit is 21.56 A
B. <span>much precipitation </span>
