Question

In a certain series RLC circuit being driven at a frequency of $\mathbf{60}\mathbf{.}\mathbf{0}\mathit{H}\mathit{z}$, the maximum voltage across the inductor is$\mathbf{2}\mathbf{.}\mathbf{00}$times the maximum voltage across the resistor and$\mathbf{2}\mathbf{.}\mathbf{00}$times the maximum voltage across the capacitor. (a) By what angle does the current lag the generator emf? (b) If the maximum generator emf is$\mathbf{30}\mathbf{.}\mathbf{0}\mathit{V}$, what should be the resistance of the circuit to obtain a maximum current of$\mathbf{300}\mathit{m}\mathit{A}$?

Short answer

a. The angle by which the current lags the generator emf is$45°$.

b. Resistance of the circuit when generator emf is $30.0V$and current is $300mA$, $70.7\Omega$.

Step-by-step solution

Step 1: Identification of the given data

The frequency is, $f=60.0 \text{Hz}$.

The maximum current is, $I=300 \text{mA}=0.3 \text{A}$

The maximum emf is, ${\epsilon }_m=30.0 \text{V}$

$V_L=2\times V_R$

$V_L=2\times V_C$

Determining the concept

The voltage across the inductor in terms of the voltage across the capacitor and resistor. Using the formula for phase constant, find the angle by which the current lags the generator emf. Find the resistance of the circuit by using Ohm’s law.

Formulae are as follows:

$\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\varphi }\mathbf{=}\left(\frac{V_L-V_C}{V_R}\right)$

$\mathit{V}\mathbf{=}\mathit{I}\mathit{R}$

Where,

Vis the potential difference.

Iis the current.

Ris the resistance.

(a) Determining the angle by which the current lags the generator emf

The phase constant is given as,

$\tan \varphi =\left(\frac{V_L-V_C}{V_R}\right)$

But,$$\begin{gathered} V_L=2\times V_R \\ {V}_R=\frac{V_L}{2} \end{gathered}$$

Also,$$\begin{gathered} V_L=2\times V_C \\ {V}_C=\frac{V_L}{2} \end{gathered}$$

Therefore,

$\begin{array}{rcl}\tan \varphi & =& \left(\frac{V_L-\left(\frac{V_L}{2}\right)}{\left(\frac{V_L}{2}\right)}\right)\\ \tan \varphi & =& \left(\frac{\frac{V_L}{2}}{\frac{V_L}{2}}\right)\\ \tan \varphi & =& 1.00\\ & & \end{array}$

$$\begin{gathered} \varphi ={\tan }^{-1}\left(1.00\right) \\ \varphi =45.0^0 \end{gathered}$$

Hence, the angle by which the current lags the generator emf is $45°$

(b) Determining the resistance of the circuit when generator emf is 30.0V and current is 300mA

By Ohm’s law,

$V=IR$

But,

$V={\epsilon }_m\cos \varphi$

Thus,

${\epsilon }_m\cos \varphi =IR$

Rearranging the terms,

$R=\frac{{\epsilon }_m\cos \varphi }{I}$

Substitute all the value in the above equation,

$\begin{array}{rcl}R& =& \frac{30.0V\times \cos 45°}{0.3A}\\ & =& 70.7\Omega \end{array}$

Hence, the resistance of the circuit when generator emf is $30.0V$and current is 300mA is$70.7\Omega$

By using the given relation between voltages across the capacitor, resistor, and inductor, and the formula for phase constant, found the angle between the voltage and current. Using Ohm’s law, found the required resistance.