Chapter 21, Problem 82QAP

To determine

The value of the rms current if the frequency of the supply is changed to 60.0 Hz.

Answer to Problem 82QAP

If the frequency of the supply is changed to 60.0 Hz, the rms current flowing in the circuit is 57.3 mA.

Explanation of Solution

Given info:

Capacitance of the capacitor

  $C=15.0\mu \text{F}=15.0\times {10}^{-6}\text{F}$

Voltage amplitude

  $V_0=25.0\text{ V}$

Resonant frequency

  $f_0=44.5\text{ Hz}$

Maximum rms current

  ${\left(i_{\text{rms}}\right)}_{\text{max}}=65.0\text{ mA}=65.0\times {10}^{-3}\text{A}$

New frequency

  $f=60.0\text{ Hz}$

Formula used:

When the current flowing in a RLC circuit connected to an ac supply is maximum, the frequency of the source is equal to Resonant frequency. When the circuit is in resonance, the impedance of the circuit is purely resistive. Therefore, the resistance R of the RLC circuit is given by,

  $R=\frac{V_{\text{rms}}}{{\left(i_{\text{rms}}\right)}_{\text{max}}}$

Where, the rms voltage is given by,

  $V_{\text{rms}}=\frac{V_0}{\sqrt{2}}$

Therefore,

  $R=\frac{V_0}{\sqrt{2}{\left(i_{\text{rms}}\right)}_{\text{max}}}$......(1)

The resonant frequency is related to the inductance L and the capacitance C as follows:

  $f_0=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}$

Rewrite the equation for L.

  $L=\frac{1}{4{\pi }^2{f}_0^{2}C}$......(2)

Using the values of L and R calculated using equations (1) and (2) the impedance Z of the circuit at a frequency f can be calculated using the expression:

  $Z=\sqrt{{\left(\frac{1}{2\pi fC}-2\pi fL\right)}^2+R^2}$......(3)

The rms current flowing in the circuit at a frequency f is given by,

  $i_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{V_0}{\sqrt{2}Z}$......(4)

Calculation:

Calculate the resistance R of the circuit by substituting the known values of the variables in equation (1).

  $\begin{array}{c}R=\frac{V_0}{\sqrt{2}{\left(i_{\text{rms}}\right)}_{\text{max}}}\\ =\frac{25.0\text{ V}}{\sqrt{2}\left(65.0\times {10}^{-3}\text{A}\right)}\\ =2.716\times {10}^2\Omega \end{array}$

Calculate the inductance of the circuit by substituting the known values of the variables in equation (2).

  $\begin{array}{c}L=\frac{1}{4{\pi }^2{f}_0^{2}C}\\ =\frac{1}{4{\left(3.14\right)}^2{\left(44.5\text{ Hz}\right)}^2\left(15.0\times {10}^{-6}\text{F}\right)}\\ =0.8536\text{ H}\end{array}$

Calculate the value of impedance of the circuit at a frequency 60.0 Hz by substituting the known values of the variables in equation (3).

  $\begin{array}{c}Z=\sqrt{{\left(\frac{1}{2\pi fC}-2\pi fL\right)}^2+R^2}\\ =\sqrt{{\left[\frac{1}{2\left(3.14\right)\left(60.0\text{ Hz}\right)\left(15.0\times {10}^{-6}\text{F}\right)}-2\left(3.14\right)\left(60.0\text{ Hz}\right)\left(0.8536\text{ H}\right)\right]}^2+{\left(2.716\times {10}^2\Omega \right)}^2}\\ =3.083\times {10}^2\Omega \end{array}$

Finally determine the rms value of current flowing in the circuit at the frequency 60.0 Hz by substituting the values of the variables in equation (4).

  $\begin{array}{c}{i}_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{V_0}{\sqrt{2}Z}\\ =\frac{25.0\text{ V}}{\sqrt{2}\left(3.083\times {10}^2\Omega \right)}\\ =5.732\times {10}^{-2}\text{A}\\ \text{=57}\text{.3 mA}\end{array}$

Conclusion:

Thus, if the frequency of the supply is changed to 60.0 Hz, the rms current flowing in the circuit is 57.3 mA.