Question

Figure 31-25 shows the currentand driving emf $\mathit{\epsilon }$for a series RLC circuit. (a) Does the current lead or lag the emf? (b) Is the circuit’s load mainly capacitive or mainly inductive? (c) Is the angular frequency $\mathbf{}{\mathit{\omega }}_{\mathbf{d}}$of the emf greater than or less than the natural angular frequency $\mathit{\omega }$?

Short answer

1. The current leads the emf.
2. The circuit’s load is mainly capacitive.
3. The angular frequency ${\omega }_d$is less than the natural angular frequency.

Step-by-step solution

The given data

Figure 31-25 showing current and driving emf for a series RLC circuit is given.

Understanding the concept of current and emf of RLC circuit

We can predict whether the current leads or lags the emf by observing the given figure. From this, we can find the type of circuit’s load and whether the angular frequency is less than the natural angular frequency. If the current is leading the emf voltage, it is considered to be capacitive and if lagging, it is considered to be inductive.

a) Calculation to predict whether the current is leading or lagging

In the given figure, we can observe that the current peaks at an earlier time than the emf.

Therefore, current leads the emf.

b) Calculation to predict whether the circuit load

Since current leads the emf, the circuit’s load is mainly capacitive.

c) Calculation to predict whether the angular frequency is less or greater than natural frequency

Since current leads the emf and phase constant is negative, the angular frequency ${\omega }_d$is less than the natural angular frequency.