Using Kirchhoff's rules, find the following. (E₁ = 70.5 V, E₂ = 61.9 V, and E3 = 78.7 V.) 4.00 ΚΩ www Rę a 2.00 ΕΩ b R₁ E, a R Σ 3.00 ΕΩ (a) the current in each resistor shown in the figure above (b) the potential difference between points c and f Step 1 We assume the currents in the three branches of this circuit are directed as shown in the diagram below. 4.00 ΚΩ www R3 d E, R₂: 13.00 ΚΩ E3 2.00 ΚΩ ww R₁ If our assumed directions for the currents are correct, the calculated currents will have positive signs. We apply Kirchhoff's junction rule at point c to obtain 12 = 1₁ + 13. [1] We apply Kirchhoff's loop rule, going clockwise around loop abcfa to obtain E₁-E₂-R₂1₂ R₁l₁ = 0. Solving for I₁, we have
We apply Kirchhoff's junction rule at point c to obtain 12 = 11 + 13. [1] We apply Kirchhoff's loop rule, going clockwise around loop abcfa to obtain E₁-E₂R₂I₂R₁I₁ = 0. Solving for I₁, we have so Solving for I3, we have so 1₁ - B₁ - B - Rol2 = (B₁ - Bo) - (R) = R₁ 1₁ = 4.3✔✔ 1.5 1₂. [2] Applying Kirchhoff's loop rule, going counterclockwise around loop edcfe, we have E3-R313-82-R₂1₂ = 0. so Submit 13 = (8₁ - 8₂ - 22₂ - (B₁-B₂) - (2²3) 2₂ 13 = R3 R3 1₂ 1₂ 4.3 x 10-3 10-³ A) - 1.5✔ 4.2 = Step 2 Substituting Equations [2] and [3] into Equation [1] from the previous step and collecting terms, we have (8.50 x 10-3 A)-([ X Your response differs from the correct answer by more than 10%. Double check your calculations.) 12 (8.50 x 10-3 4.2 x 10-3 Skip (you cannot come back) A) - (0.75 ✔ 0.75 mA. [3]
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✔ 1.5 1₂. [2] Applying Kirchhoff's loop rule, going counterclockwise around loop edcfe, we have E3-R313-82-R₂1₂ = 0. so Submit 13 = (8₁ - 8₂ - 22₂ - (B₁-B₂) - (2²3) 2₂ 13 = R3 R3 1₂ 1₂ 4.3 x 10-3 10-³ A) - 1.5✔ 4.2 = Step 2 Substituting Equations [2] and [3] into Equation [1] from the previous step and collecting terms, we have (8.50 x 10-3 A)-([ X Your response differs from the correct answer by more than 10%. Double check your calculations.) 12 (8.50 x 10-3 4.2 x 10-3 Skip (you cannot come back) A) - (0.75 ✔ 0.75 mA. [3]