91Ó°ÊÓ

The values of phase constantare

1. $\mathrm{ϕ}=-15°$
   
2. $\mathrm{ϕ}=+35°$
   
3. $\mathrm{Ï•}=\frac{\mathrm{Ï€}}{3}\text{rad}$
   
4. $\mathrm{Ï•}=-\frac{\mathrm{Ï€}}{6}\text{rad}$
   

Alternating current (ac) is the flow of electric charge that periodically reverses direction. The current produces an alternating emf in a circuit. Using Eq.31-28 and Eq.31-29 and substituting the given values of phase constant, we can find in which values the load is primarily capacitive and in which the current leads the alternating emf. The AC lags behind the emf when an ac passes through the inductor.

Formulae:

The alternating emf produced in the circuit,$\mathit{Î\mu }\mathbf{=}{\mathit{Î\mu }}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{\left(}{\mathbf{Ó\neg }}_{\mathbf{d}}\mathbf{t}\mathbf{\right)}\mathbf{}$
(i)

The current driven in the circuit, $\mathit{i}\mathbf{=}\mathit{I}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}{\mathbf{Ó\neg }}_{\mathbf{d}}\mathbf{t}\mathbf{-}\mathbf{Ï•}\mathbf{)}$(ii)

For phase constant,$\mathrm{ϕ}=-15°$:

From equations (i) and (ii), phase constant $\mathrm{ϕ}=-15°$, this is negative. Thus, the current leads alternating emf. Therefore, we can write that the load is primarily capacitive.

For phase constant $\mathrm{ϕ}=+35°$,

From equations (i) and (ii), phase constant $\mathrm{ϕ}=+35°$, this is positive. Thus, the current lags alternating emf. Therefore, we can write that the load is primarily inductive.

For phase constant $\mathrm{Ï•}=\frac{\mathrm{Ï€}}{3}rad$:

From equations (i) and (ii), phase constant$\mathrm{Ï•}=\frac{\mathrm{Ï€}}{3}rad$,this is positive. Thus, the current lags alternating emf. Therefore, we can write that the load is primarily inductive.

For phase constant $\mathrm{ϕ}=-15°$:

From Eq.31-28 and Eq.31-29, phase constant $\mathrm{Ï•}=-\frac{\mathrm{Ï€}}{6}\text{rad},$which is negative. Thus, the current leads alternating emf. Therefore, we can write that the load is primarily capacitive.

Hence, in phase constant $\mathrm{ϕ}=-15°$ and $\mathrm{ϕ}=-\frac{\mathrm{π}}{6}\text{rad},$the load is primarily capacitive.

From the solution and calculations of part (a), we have that in (2) and (3), the current lags the alternating emf.