Let's solve the problems step by step: --- ### **24. What is the current in the 12 Ω resistor immediately after switch S is thrown?** **Step 1: Understand the circuit when switch S is just closed.** - The inductor (2 H) initially opposes any change in current through it, so it acts like an **open circuit** immediately after the switch is thrown. **Step 2: Analyze the circuit with the inductor as an open circuit.** - The branch containing the inductor and the 12 Ω resistor will have **no current** immediately after S is closed. - The current will only flow through the top branch (the two 6 Ω resistors in series). **Step 3: Calculate the current in the 12 Ω resistor.** - Since the inductor behaves as an open circuit, current through 12 Ω resistor = **0 A**. **Final Answer:** **A. 0** --- ### **25. After S has been closed for a long time, it is opened again. How much energy will be consumed in the circuit resistors after S is opened?** **Step 1: When S is closed for a long time, the inductor behaves as a short circuit (inductor is just a wire for DC steady state).** - The total resistance in the circuit is: - The two 6 Ω resistors in series: \(6 + 6 = 12 \Omega\) - This 12 Ω is in parallel with the 12 Ω resistor. - Equivalent resistance of two 12 Ω resistors in parallel: \[ R_{eq} = \frac{12 \times 12}{12 + 12} = \frac{144}{24} = 6\ \Omega \] **Step 2: Calculate the total current from the battery.** \[ I_{total} = \frac{V}{R_{eq}} = \frac{90}{6} = 15\ \text{A} \] **Step 3: Current splits between the two parallel branches:** - Since both branches have 12 Ω resistance, the current splits equally: - Each branch gets \(7.5\) A. - The current through the inductor (bottom branch) is **7.5 A** just before S is opened. **Step 4: Energy stored in the inductor** - The energy stored in an inductor is: \[ E = \frac{1}{2} L I^2 \] Where \(L = 2\) H, \(I = 7.5\) A: \[ E = \frac{1}{2} \times 2 \times (7.5)^2 = 1 \times 56.25 = 56.25\ \text{J} \] **Step 5: When the switch is opened, all this stored energy is dissipated in the resistors of the branch containing the inductor and 12 Ω resistor.** **Final Answer:** None of the options matches 56.25 J exactly, but the closest value is: **D. 36 J** But this is not the correct calculated value. The answer should be **56.25 J** based on the calculation above. If there's a typo in the options, choose the closest, which is **36 J**, but the correct physics answer is **56.25 J**. --- #### **Summary:** - 24: **A. 0** - 25: **56.25 J (closest option D. 36 J, but see explanation above)** If you have to choose from the provided options, pick **A for 24** and **D for 25**, but note the calculation explanation!