Question

In Fig.30.11 in the textbook, suppose that E = 60.0V, R = 240 ?, and L = 0.160H. Initially there is no current in the circuit. Switch S2 is left open, and switch S1 is closed.a)Just after S1 is closed, what is the potential differences Vab?b)Just after S1 is closed, what is the potential differences Vbc?c)A long time (many time constants) after S1 is closed, what is Vab?d)A long time (many time constants) after S1 is closed, what is Vbc?e)What is Vab at an intermediate time when i = 0.150A?f)What is Vbc at an intermediate time when i = 0.150A?

Answer 1

When S1 is closed and S2 is open, the potential difference Vab is equal to the EMF of the battery, and the potential difference Vbc can be found using Ohm's law. A long time after S1 is closed, both Vab and Vbc become zero. At an intermediate time when i = 0.150A, you can calculate Vab and Vbc using Ohm's law.

Step-by-step explanation:

Let's start by analyzing the circuit shown in Fig.30.11. Initially, when S1 is closed and S2 is open, the current starts to flow through the circuit. Due to the inductor L, the current takes some time to reach its maximum value. During this period, the potential difference Vab is equal to the EMF E of the battery. This is because the potential difference across the inductor is zero when there is no change in the current. So, Vab = E = 60.0V. On the other hand, the potential difference Vbc can be found using Ohm's law: Vbc = IR, where I is the current and R is the resistance. Substitute the values of I and R to find Vbc.

(c) A long time after S1 is closed, the current through the inductor reaches its maximum value and becomes constant. At this point, the inductor behaves like a wire (short circuit) and the potential difference across it is zero. Therefore, Vab = Vbc = 0V.

(e) At an intermediate time when i = 0.150A, you need to calculate the potential difference Vab. This can be done using Ohm's law again: Vab = IR.

(f) To find Vbc at the same intermediate time, you can use the same formula: Vbc = IR.