Question

Figure 31-23 shows the current i and driving emffor a series RLC circuit. (a) Is the phase constant positive or negative? (b) To increase the rate at which energy is transferred to the resistive load, should L be increased or decreased? (c) Should, instead, C be increased or decreased?

Short answer

1. The phase constant is positive.
2. To increase the rate, at which energy is transferred to the resistive load,$L$should be decreased.
3. The capacitance $C$should be decreased.

Step-by-step solution

The given data

1. The current is$i$
2. Driving emf is $\epsilon$.
3. Fig.31-23 having a series RLC circuit is shown.

Understanding the concept of the RLC circuit

Using Eq.31-28 and Eq.31-29, we can find whether the phase constant is positive or negative. Using Eq.31-76 and from the vector diagram of emf and current, we can find whether should be increased or decreased to increase the rate at which energy is transferred to the resistive load, and also predicts whether the capacitance value should be increased or decreased instead.

Formulae:

From Eq.31-28, the equation of the alternating emf is given by, $\mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{\left(}{\mathbf{\omega }}_{\mathbf{d}}\mathbf{t}\mathbf{\right)}\mathbf{}$ (i)

The current is driven in the circuit, $\mathit{i}\mathbf{=}\mathit{I}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}{\mathbf{\omega }}_{\mathbf{d}}\mathbf{t}\mathbf{-}\mathbf{\varphi }\mathbf{)}$ (ii)

The average power of the circuit, ${\mathit{P}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{r}\mathbf{m}\mathbf{s}}{\mathit{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }$ (iii)

where,${\mathit{\epsilon }}_{\mathbf{r}\mathbf{m}\mathbf{s}}$and${\mathit{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}$are constant.

The voltage equation from Ohm’s law, $\mathit{V}\mathbf{=}\mathit{I}\mathit{R}\mathbf{ }{\mathbf{\text{or IX}}}_{\mathbf{c}}{\mathbf{\text{or IX}}}_{\mathbf{L}}$(iv)

a) Calculation of the phase constant

From equation (i), we can see that the $i$curve is shifted to the right. Therefore, the phase constant$\varphi$is positive. Thus, from equations (i) and (ii), we can write that the current $i$lags emf $\epsilon$. Here, the current is inductive.

b) Calculation of the inductance behavior to increase the energy rate

From equation (iii), we can see that to increase the value of $P_{avg}$, we have to increase $cos\varphi$. That will let us reduce $\varphi$.

From above figure, we get

$$\begin{gathered} tan\varphi =\frac{\left(V_L-V_C\right)}{V_R} \\ =\frac{\left(X_L-X_C\right)}{R}.................\left(a\right) \end{gathered}$$

Since,$V_L-V_C>0$. Therefore,$\varphi$must be positive.

To reduce $\varphi$, we have to reduce$tan\varphi$and therefore,$X_L$also decrease.

But, the reactance of the inductance,$X_L={\omega }_{d}L$

Hence, to reduce$X_L$, inductance $L$is decreased.

c) Calculation of the capacitance behavior

To reduce $\varphi$, we have to reduce$tan\varphi$

Therefore, increase the term of capacitance reactance of equation (a) as follows:

$X_C=\frac{1}{{\omega }_{d}C}$

Hence, as $X_C$increases then $C$decreases.