Chapter 21, Problem 69QAP

To determine

(a)

The impedance of a LRC circuit with 500 O resistor, 0.20 H inductor and 2.0 μF capacitor connected to a 100 V peak, 500 Hz ac source.

Answer to Problem 69QAP

The impedance of a LRC circuit with 500 O resistor, 0.20 H inductor and 2.0 μF capacitor connected to a 100 V peak, 500 Hz ac source is found to be 690 Ω.

Explanation of Solution

Given:

$\begin{array}{c}L=0.20\text{ H}\\ C=2.0\mu \text{F}=2.0\times {10}^{-6}\text{F}\\ R=500\Omega \\ f=\text{500 Hz}\\ V_{\text{0}}=100\text{ V}\end{array}$

Formula used:

The impedance Z of an ac circuit is given by the expression,

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

Calculation:

Substitute the values of the variables in the formula and calculate the value of impedance of the circuit.

  $\begin{array}{c}Z=\sqrt{{\left(\frac{1}{2\pi fC}-2\pi fL\right)}^2+R^2}\\ =\sqrt{{\left(\frac{1}{2\left(3.14\right)\left(\text{500 Hz}\right)\left(2.0\times {10}^{-6}\text{F}\right)}-2\left(3.14\right)\left(\text{500 Hz}\right)\left(0.20\text{ H}\right)\right)}^2+{\left(500\Omega \right)}^2}\\ =685.5\Omega \\ =690\Omega \end{array}$

Conclusion:

Thus, the impedance of a LRC circuit with 500 O resistor, 0.20 H inductor and 2.0 μF capacitor connected to a 100 V peak, 500 Hz ac source is found to be 690 Ω.

To determine

(b)

The amplitude of the current from the source.

Answer to Problem 69QAP

The amplitude of the current from the source is found to be 0.15 A.

Explanation of Solution

Given:

  $\begin{array}{c}{V}_{\text{0}}=100\text{ V}\\ Z=685.5\Omega \end{array}$

Formula used:

The amplitude of the current from the source is equal to the peak value of current. This is given by,

  $i_{\text{0}}=\frac{V_{\text{0}}}{Z}$

Calculation:

Substitute the values of the variables in the formula and calculate the value of the peak current.

  $\begin{array}{c}{i}_{\text{0}}=\frac{V_{\text{0}}}{Z}\\ =\frac{100\text{ V}}{685.5\Omega \Omega }\\ =0.146\text{ A}\\ =0.\text{15 A}\end{array}$

Conclusion;Thus, the amplitude of the current from the source is found to be 0.15 A.

To determine

(c)

The equation which shows the variation of current with time.

Answer to Problem 69QAP

The current in the given circuit obeys the following equation:

  $i\left(t\right)=\left(0.15\text{ A}\right)\sin \left(1000\pi t-0.754\text{ rad}\right)$.

Explanation of Solution

Given:

  $\begin{array}{c}L=0.20\text{ H}\\ C=2.0\mu \text{F}=2.0\times {10}^{-6}\text{F}\\ R=500\Omega \\ f=\text{500 Hz}\\ i_0=0.15\text{ A}\end{array}$

The equation showing the variation of voltage with time

  $V\left(t\right)=\left(100\text{ V}\right)\sin \left(1000\pi t\right)$

Formula used:

The current in the circuit varies with time according to the equation:

  $i\left(t\right)=i_0\sin \left(1000\pi t+\varphi \right)$......(1)

The phase angle is given by the expression,

  $\varphi =\arctan \left[\frac{1}{R}\left(\frac{1}{2\pi fC}-2\pi fL\right)\right]$......(2)

Calculation:

Calculate the phase angle between current and voltage by substituting the values of the variables in equation (2).

  $\begin{array}{c}\varphi =\arctan \left[\frac{1}{R}\left(\frac{1}{2\pi fC}-2\pi fL\right)\right]\\ =\arctan \left[\frac{1}{\left(500\Omega \right)}\left(\frac{1}{2\left(3.14\right)\left(\text{500 Hz}\right)\left(2.0\times {10}^{-6}\text{F}\right)}-2\left(3.14\right)\left(\text{500 Hz}\right)\left(0.20\text{ H}\right)\right)\right]\\ =\arctan \left(\frac{-469\Omega }{500\Omega }\right)\\ =-43.2°\end{array}$

Express the angle in radians.

  $\begin{array}{c}\varphi =\left(-43.2°\right)\left(\frac{\pi \text{ rad}}{180°}\right)\\ =\frac{-43.2°\times 3.14\text{ rad}}{180°}\\ =-0.754\text{ rad}\end{array}$

Substitute the value of $i_0$ and $\varphi$ in equation (1).

  $\begin{array}{c}i\left(t\right)=i_0\sin \left(1000\pi t+\varphi \right)\\ =\left(0.15\text{ A}\right)\sin \left(1000\pi t-0.754\text{ rad}\right)\end{array}$

Conclusion:

Thus the current in the given circuit obeys the following equation:

  $i\left(t\right)=\left(0.15\text{ A}\right)\sin \left(1000\pi t-0.754\text{ rad}\right)$.