Question

In a circuit with five resistors and two batteries, what are the currents I1, I2, I3, and I4? (Consider the following values: R1 = 1.16 ohm, R2 = 2.120 ohm, R3 = 3.09 ohm, R4 = 4.200 ohm, R5 = 6.160 ohm. Due to the nature of this problem, do not use rounded intermediate values in your calculations.)

Option 1: I1
Option 2: I2
Option 3: I3
Option 4: I4

Answer 1

Option 1: I₁ = (6.16V - 4.20V) / (1.16Ω + 2.12Ω)

Option 2: I₂ = (6.16V - 3.09V) / (1.16Ω + 2.12Ω + 3.09Ω)

Option 3: I₃ = (6.16V - 2.120V) / (1.16Ω + 2.12Ω + 3.09Ω + 4.20Ω)

Option 4: I₄ = (6.16V - 1.16V) / (1.16Ω + 2.12Ω + 3.09Ω + 4.20Ω + 6.16Ω)

Step-by-step explanation:

To find the currents I₁, I₂, I₃, and I₄ in the circuit with resistors R₁, R₂, R₃, R₄, and R₅, and two batteries, we use Ohm's Law and Kirchhoff's laws.

For I₁:

I₁ = (V_battery1 - V_R₁) / (R₁ + R₂)

For I₂:

I₂ = (V_battery1 - V_R₁ - V_R₂) / (R₁ + R₂ + R₃)

For I₃:

I₃ = (V_battery1 - V_R₁ - V_R₂ - V_R₃) / (R₁ + R₂ + R₃ + R₄)

For I₄:

I₄ = (V_battery1 - V_R₁ - V_R₂ - V_R₃ - V_R₄) / (R₁ + R₂ + R₃ + R₄ + R₅)

These equations give us the currents through each resistor in the circuit.