This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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To determine the current in each branch of the circuit in FIGURE 1 using Kirchhoff's laws, we will apply Kirchhoff's Voltage Law (KVL) to the two independent loops (meshes) and solve for the mesh currents I_1 and I_2. The directions of I_1 and I_2 are given as clockwise in the diagram. Step 1: Apply KVL to Mesh 1 (left loop, AFEBA) We sum the voltage drops around the loop in the clockwise direction. The current I_1 flows through the 1 \, resistor. The current in the common branch BE is (I_1 - I_2) flowing from B to E. The voltage sources are E_1 = 15 \, V (positive at A, negative at F) and E_2 = 12 \, V (positive at B, negative at E). Starting from point A and moving clockwise: Voltage drop across 1 \, resistor (AF): I_1 × 1 \, . Voltage source E_1: Moving from A to F, we go from the positive terminal to the negative terminal of E_1. This is a voltage drop, so -15 \, V. Voltage drop across 0.2 \, resistor (BE): We are moving from E to B. The current (I_1 - I_2) flows from B to E. So, moving from E to B, we are going against (I_1 - I_2). This results in a voltage rise, so -(I_1 - I_2) × 0.2 \, . Voltage source E_2: Moving from E to B, we go from the negative terminal to the positive terminal of E_2. This is a voltage rise, so +12 \, V. Summing these voltages to zero: I_1(1) - 15 - (I_1 - I_2)(0.2) + 12 = 0 I_1 - 15 - 0.2I_1 + 0.2I_2 + 12 = 0 0.8I_1 + 0.2I_2 - 3 = 0 0.8I_1 + 0.2I_2 = 3 Multiplying by 10 to clear decimals: 8I_1 + 2I_2 = 30 Dividing by 2: 4I_1 + I_2 = 15 (Equation 1) Step 2: Apply KVL to Mesh 2 (right loop, BCDEB) We sum the voltage drops around the loop in the clockwise direction. The current I_2 flows through the 0.2 \, resistor in branch CD. The current in the common branch BE is (I_1 - I_2) flowing from B to E. The voltage sources are E_2 = 12 \, V (positive at B, negative at E) and E_3 = 9 \, V (positive at C, negative at D). Starting from point B and moving clockwise: Voltage drop across 0.2 \, resistor (BE): We are moving from B to E. The current (I_1 - I_2) flows from B to E. This is a voltage drop, so (I_1 - I_2) × 0.2 \, . Voltage source E_2: Moving from B to E, we go from the positive terminal to the negative terminal of E_2. This is a voltage drop, so -12 \, V. Voltage drop across 0.2 \, resistor (CD): I_2 × 0.2 \, . Voltage source E_3: Moving from C to D, we go from the positive terminal to the negative terminal of E_3. This is a voltage drop, so -9 \, V. Summing these voltages to zero: (I_1 - I_2)(0.2) - 12 + I_2(0.2) - 9 = 0 0.2I_1 - 0.2I_2 - 12 + 0.2I_2 - 9 = 0 0.2I_1 - 21 = 0 0.2I_1 = 21 I_1 = (21)/(0.2) = 105 \, A (Equation 2) Step 3: Solve the system of equations We already found I_1 from Equation 2. Now substitute I_1 = 105 \, A into Equation 1: 4(105) + I_2 = 15 420 + I_2 = 15 I_2 = 15 - 420 I_2 = -405 \, A Step 4: Determine the current in each branch The currents in the branches are: Current in branch AF (and FE, ED): This is the mesh current I_1. I_AF = I_1 = 105 A Current in branch BE: This is the difference between the mesh currents, (I_1 - I_2), flowing from B to E. I_BE = I_1 - I_2 = 105 \, A - (-405 \, A) = 105 \, A + 405 \, A = 510 A Current in branch CD (and BC): This is the mesh current I_2. I_CD = I_2 = -405 A The negative sign indicates that the actual current flows in the opposite direction to the assumed clockwise direction for I_2, i.e., from D to C. What's next?