Question

While the instantaneous electric power delivered by a single-phase generator under balanced steady-state conditions is a function of time having two components of a constant and a double-frequency sinusoid, the total instantaneous electric power delivered by a three-phase generator under balanced steady-state conditions is a constant.

(a) True

(b) False

Short answer

Answer

Therefore the correct option is (a): True

Step-by-step solution

Expression for the power in case of single phase generator

The instantaneous power in case of single-phase generator under balanced steady-state condition is given by;

$$\begin{gathered} {\mathbf{P}}_{\mathbf{an}}\left(t\right)\mathbf{=}{\mathbf{V}}_{\mathbf{an}}\left(t\right){\mathbf{i}}_{\mathbf{a}}\left(t\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\left(\sqrt{2}{V}_{\mathrm{LN}}\cos \left(\mathrm{\omega t}+{\theta }_1\right)\right)\mathbf{}\left(\sqrt{2}{I}_L\cos \left(\mathrm{\omega t}+{\theta }_2\right)\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}{\mathbf{V}}_{\mathbf{LN}}\mathbf{cos}\left(\mathrm{\omega t}+{\theta }_1\right)\mathbf{}{\mathbf{I}}_{\mathbf{L}}\mathbf{cos}\left(\mathrm{\omega t}+{\theta }_2\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{V}}_{\mathbf{LN}}{\mathbf{I}}_{\mathbf{L}}\mathbf{cos}\left({\theta }_1+{\theta }_2\right)\mathbf{+}{\mathbf{V}}_{\mathbf{LN}}\mathbf{}{\mathbf{I}}_{\mathbf{L}}\mathbf{cos}\mathbf{}\left(2\mathrm{\omega t}+{\theta }_1+{\theta }_2\right) \end{gathered}$$

Therefore, the expression for the power in case of single-phase generator is function of time.

Expression for the power in case of three- phase generator

Consider the expression for the power in case of three-phase generator

$$\begin{gathered} P_{3\varnothing }\left(t\right)=P_{3\varnothing } \\ =3V_{NL}{I}_L\cos \left({\theta }_1-{\theta }_2\right) \end{gathered}$$

So the expression for the power in case of three-phase generator is not a function of time.

Thus, the correct option is (a).