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 solve this circuit, we will use nodal analysis. Step 1: Define nodes and assign a reference node. Let the top horizontal line be Node A with potential V_A. Let the bottom horizontal line be Node B with potential V_B. We set Node B as the reference node, so V_B = 0V. Step 2: Apply Kirchhoff's Current Law (KCL) at Node B. Based on the current arrows in the diagram: • Current I_3 enters Node B. • Current I_1 leaves Node B. • Current I_2 leaves Node B. Therefore, KCL at Node B states that the sum of currents entering the node equals the sum of currents leaving the node: I_3 = I_1 + I_2 (Equation 1) Step 3: Express branch currents in terms of node voltages. We will write expressions for I_1, I_2, and I_3 using Ohm's Law and considering the voltage sources in each branch. The current direction is from higher potential to lower potential. Since the arrows for I_1, I_2, I_3 point from Node A towards Node B (or through components to Node B), we assume V_A is higher than V_B. For the leftmost branch (containing the 2 resistor, 15V source, and 3 resistor): Current I_3 flows from Node A, through the 2 resistor, then through the 15V source (positive terminal up, negative down), then through the 3 resistor, and finally to Node B. The voltage drop across the resistors and the source in the direction of I_3 is: V_A - I_3(2) - 15V - I_3(3) = V_B Since V_B = 0V: V_A - 5I_3 - 15 = 0 I_3 = (V_A - 15)/(5) (Equation 2) For the middle branch (containing the 4 resistor and 10V source): Current I_2 flows from Node A, through the 4 resistor, then through the 10V source (positive terminal up, negative down), and finally to Node B. V_A - I_2(4) - 10V = V_B Since V_B = 0V: V_A - 4I_2 - 10 = 0 I_2 = (V_A - 10)/(4) (Equation 3) For the rightmost branch (containing the 5 resistor and 5V source): Current I_1 flows from Node A, through the 5 resistor, then through the 5V source (positive terminal up, negative down), and finally to Node B. V_A - I_1(5) - 5V = V_B Since V_B = 0V: V_A - 5I_1 - 5 = 0 I_1 = (V_A - 5)/(5) (Equation 4) Step 4: Substitute into KCL and solve for V_A. Substitute Equations 2, 3, and 4 into Equation 1: (V_A - 15)/(5) = (V_A - 5)/(5) + (V_A - 10)/(4) To eliminate the denominators, multiply the entire equation by the least common multiple of 5 and 4, which is 20: 4(V_A - 15) = 4(V_A - 5) + 5(V_A - 10) 4V_A - 60 = 4V_A - 20 + 5V_A - 50 4V_A - 60 = 9V_A - 70 Rearrange the terms to solve for V_A: 70 - 60 = 9V_A - 4V_A 10 = 5V_A V_A = (10)/(5) = 2V Step 5: Calculate the currents I_1, I_2, I_3. Now substitute V_A = 2V back into Equations 2, 3, and 4: For I_3: I_3 = (2V - 15V)/(5) = (-13V)/(5) = -2.6A For I_2: I_2 = (2V - 10V)/(4) = (-8V)/(4) = -2A For I_1: I_1 = (2V - 5V)/(5) = (-3V)/(5) = -0.6A The negative signs indicate that the actual direction of current flow is opposite to the direction indicated by the arrows in the diagram. The final answers are: I_1 = -0.6A I_2 = -2A I_3 = -2.6A Send me the next one 📸