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 calculate the current and voltage drop across the 11.2 \, resistor using Norton's theorem, we follow these steps: Step 1: Find the Norton Current (I_N) To find I_N, we remove the load resistor (R_load = 11.2 \, ) and place a short circuit across its terminals. Let the terminals be B and ground. This means the voltage at terminal B is 0 \, V. The circuit now consists of the 20 \, V source, R_1 = 20 \, , R_2 = 12 \, , and R_3 = 8 \, . Let Node A be the junction between R_1, R_2, and R_3. When terminal B is shorted to ground, R_2 is connected between Node A and ground, and R_3 is also connected between Node A and ground. Thus, R_2 and R_3 are in parallel. The equivalent resistance of R_2 R_3 is: R_23 = (R_2 × R_3)/(R_2 + R_3) = (12 \, × 8 \, )/(12 \, + 8 \, ) = (96 \, ^2)/(20 \, ) = 4.8 \, The total current (I_total) flowing from the source is through R_1 and then through the parallel combination of R_2 and R_3: I_total = (V)/(R_1 + R_23) = 20 \, V20 \, + 4.8 \, = 20 \, V24.8 \, = (200)/(248) \, A = (25)/(31) \, A This current I_total reaches Node A and splits between R_2 and R_3. The Norton current I_N is the current flowing through the short circuit, which is the current that would flow through R_2 in this configuration (since the short is across the terminals where R_load was, and R_2 is connected to that node). Using the current divider rule: I_N = I_total × (R_3)/(R_2 + R_3) = (25)/(31) \, A × (8 \, )/(12 \, + 8 \, ) = (25)/(31) \, A × (8)/(20) = (25)/(31) \, A × (2)/(5) = (10)/(31) \, A I_N ≈ 0.3226 \, A Step 2: Find the Norton Resistance (R_N) To find R_N, we turn off the independent voltage source (replace it with a short circuit) and look into the terminals where the load resistor was connected (terminals B and ground). With the voltage source shorted, R_1 is connected from Node A to ground. R_3 is also connected from Node A to ground. So, R_1 and R_3 are in parallel. R_13 = (R_1 × R_3)/(R_1 + R_3) = (20 \, × 8 \, )/(20 \, + 8 \, ) = (160 \, ^2)/(28 \, ) = (40)/(7) \, Now, looking from terminals B to ground, R_2 is in series with the parallel combination of R_1 and R_3. R_N = R_2 + R_13 = 12 \, + (40)/(7) \, = (12 × 7 + 40)/(7) \, = (84 + 40)/(7) \, = (124)/(7) \, R_N ≈ 17.7143 \, Step 3: Calculate the current through the load resistor (I_load) Now we connect the load resistor R_load = 11.2 \, in parallel with the Norton equivalent circuit (I_N in parallel with R_N). We use the current divider rule: I_load = I_N × (R_N)/(R_N + R_load) I_load = (10)/(31) \, A × (124)/(7) \, (124)/(7) \, + 11.2 \, Convert 11.2 \, to a fraction: 11.2 = (112)/(10) = (56)/(5). I_load = (10)/(31) \, A × (124)/(7)(124)/(7) + (56)/(5) = (10)/(31) \, A × (124)/(7)(124 × 5 + 56 × 7)/(35) I_load = (10)/(31) \, A × (124)/(7)(620 + 392)/(35) = (10)/(31) \, A × (124)/(7)(1012)/(35) I_load = (10)/(31) \, A × (124)/(7) × (35)/(1012) = (10)/(31) \, A × (124 × 5)/(1012) I_load = (6200)/(31372) \, A = (1550)/(7843) \, A I_load ≈ 0.1976 A Step 4: Calculate the voltage drop across the load resistor (V_load) Using Ohm's Law: V_load = I_load × R_load V_load = (1550)/(7843) \, A × 11.2 \, = (1550)/(7843) \, A × (56)/(5) \, V_load = (310 × 56)/(7843) \, V = (17360)/(7843) \, V V_load ≈ 2.213 V Send me the next one 📸