Question

An alternating source drives a series $\mathit{R}\mathit{L}\mathit{C}\mathbf{}$circuit with emf amplitude of, $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{V}}$at a phase angle of$\mathbf{+}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$. When the potential difference across the capacitor reaches its maximum positive value of$\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{V}}$, what is the potential difference across the inductor (sign included)?

Short answer

The potential difference across the inductor is $-8.00\text{V}$.

Step-by-step solution

The given data

1. Emf amplitude${\epsilon }_m=6.00\text{V}$
2. Phase angle,$\varphi =30°$
3. Potential difference across capacitor$V_C=5.00\text{V}$

Understanding the concept of phasor diagram

Using the phasor diagram and applying the Pythagorean Theorem, we can find the potential difference across the inductor. The theorem applies to the right-angled triangle relating hypotenuse, perpendicular, and base of the triangle.

Calculation of the potential difference

We have the following phasor diagram:

From the phasor diagram, we can apply the Pythagorean Theorem considering equation (i) to the diagram as follows:

$$\begin{gathered} {\epsilon }_m^2=V_R^2+{\left(V_L-V_C\right)}^2 \\ {V}_L=V_c+\sqrt{{\epsilon }_m^2-V_R^2} \\ =V_c+\sqrt{{\epsilon }_m^2-{\epsilon }_m^2{\cos }^2\varphi }\text{(}\because V_R={\epsilon }_{m}cos\varphi ,\text{from diagram}) \\ =V_c+{\epsilon }_{m}sin\varphi \end{gathered}$$

For the given values, we get

$$\begin{gathered} V_L=5.00\text{V}+\left(6.00\text{V}\right)\text{sin}30° \\ =8.00\text{V} \end{gathered}$$

But $V_L$and $V_C$are $180°$out of the phase.

Therefore, the value of the potential difference across the inductor is $-8.00\text{V}$.