91Ó°ÊÓ

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

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.

We have the following phasor diagram:

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

$$\begin{gathered} {\mathrm{Î\mu }}_m^2=V_R^2+{\left(V_L-V_C\right)}^2 \\ {V}_L=V_c+\sqrt{{\mathrm{Î\mu }}_m^2-V_R^2} \\ =V_c+\sqrt{{\mathrm{Î\mu }}_m^2-{\mathrm{Î\mu }}_m^2{\cos }^2\mathrm{Ï•}}\text{(}âˆ\mu V_R={\mathrm{Î\mu }}_{m}cos\mathrm{Ï•},\text{from diagram}) \\ =V_c+{\mathrm{Î\mu }}_{m}sin\mathrm{Ï•} \end{gathered}$$

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 $-8.00\text{V}$.