This problem describes an R-L-C series circuit, and the task is to find the voltage across the inductor. Let's break it down step by step.

Given:

• Voltage source (V): 100 V
• Resistance (R): 40 Ω
• Inductive reactance (XL): 150 Ω
• Capacitive reactance (XC): 120 Ω

To find the potential difference across the inductor, we first need to calculate the total impedance of the circuit and the current flowing through it.

Step 1: Calculate the net reactance (X)

The net reactance is the difference between the inductive and capacitive reactance: $X = X_L - X_C = 150 \, \Omega - 120 \, \Omega = 30 \, \Omega$

Step 2: Calculate the total impedance (Z)

The total impedance of the series circuit is given by: $Z = \sqrt{R^2 + X^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \Omega$

Step 3: Calculate the current (I)

Using Ohm's law, the current in the circuit is: $I = \frac{V}{Z} = \frac{100 \, V}{50 \, \Omega} = 2 \, A$

Step 4: Calculate the voltage across the inductor

The voltage across the inductor can be calculated using the current and the inductive reactance: $V_L = I \times X_L = 2 \, A \times 150 \, \Omega = 300 \, V$

Answer:

The voltage across the inductor is 300 V, so the correct option is c. 300 V.

Would you like more details on any part of the explanation?

Here are 5 related questions:

1. How is the net reactance in an R-L-C circuit calculated?
2. What is the significance of impedance in AC circuits?
3. How would the result change if the reactances $X_L$ and $X_C$ were equal?
4. What happens to the current if only the resistance is increased in this circuit?
5. How is power consumption affected by reactance in an R-L-C circuit?

Tip: In an R-L-C series circuit, impedance determines both the current and how voltages distribute across individual components.