Chapter 21, Problem 81QAP

To determine

(a)

The impedance of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply.

Answer to Problem 81QAP

The impedance of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 2.18 O.

Explanation of Solution

Given:

Number of windings in the coil

  $N=750\text{ turns}$

Length of the coil

  $l=8.50\text{ cm}=8.50\times {10}^{-2}\text{m}$

Diameter of the coil

  $d=1.25\text{ cm}=1.25\times {10}^{-2}\text{m}$

Resistance of the coil

  $R=2.15\Omega$

Capacitance of the capacitor

  $C=2240\mu \text{F}=2.24\times {10}^{-3}\text{F}$

Frequency of the supply

  $f=60.0\text{ Hz}$

The rms voltage of the supply

  $V_{\text{rms}}=120\text{ V}$

Formula used:

The heating coil behaves as an inductor. The inductance L of the coil is given by,

  $L=\frac{{\mu }_0{N}^{2}A}{l}$

Here, ${\mu }_0$ is the permeability of free space and has a value ${\mu }_0=4\pi \times {10}^{-7}\text{H/m}$, A is the area of the coil, given by,

  $A=\pi r^2=\pi {\left(\frac{d}{2}\right)}^2$

And r is the radius of the coil.

Therefore, the inductance is given by,

  $\begin{array}{c}L=\frac{\left(4\pi \times {10}^{-7}\text{H/m}\right)N^2\pi {\left(\frac{d}{2}\right)}^2}{l}\\ =\frac{\left({\pi }^2\times {10}^{-7}\text{H/m}\right)N^2{d}^2}{l}\end{array}$......(1)

The impedance of the circuit is given by,

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

Calculation:

Determine the value of inductance of the heating coil by substituting the values of variables in equation (1).

  $\begin{array}{c}L=\frac{\left({\pi }^2\times {10}^{-7}\text{H/m}\right)N^2{d}^2}{l}\\ =\frac{{\left(3.14\right)}^2\left({10}^{-7}\text{ H/m}\right){\left(750\text{ turns}\right)}^2{\left(1.25\times {10}^{-2}\text{m}\right)}^2}{8.50\times {10}^{-2}\text{m}}\\ =1.019\times {10}^{-3}\text{H}\end{array}$

Substitute the value of the known variables in equation (2) and calculate the value of impedance.

  $\begin{array}{c}Z=\sqrt{{\left(\frac{1}{2\pi fC}-2\pi fL\right)}^2+R^2}\\ =\sqrt{{\left[\frac{1}{2\left(3.14\right)\left(60.0\text{ Hz}\right)\left(2.24\times {10}^{-3}\text{F}\right)}-2\left(3.14\right)\left(60.0\text{ Hz}\right)\left(1.019\times {10}^{-3}\text{H}\right)\right]}^2+{\left(2.15\Omega \right)}^2}\\ =2.184\Omega \end{array}$

Conclusion:

Thus, the impedance of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 2.18 O.

To determine

(b)

The rms current through of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply.

Answer to Problem 81QAP

The rms current through of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 55 A.

Explanation of Solution

Given:

The rms voltage of the supply

  $V_{\text{rms}}=120\text{ V}$

Impedance

  $Z=2.184\Omega$

Formula used:

The rms current flowing in the circuit is given by,

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

Calculation:

Calculate the rms value of the current by substituting the given values of variables in the formula.

  $\begin{array}{c}{i}_{\text{rms}}=\frac{V_{\text{rms}}}{Z}\\ =\frac{120\text{ V}}{2.184\Omega }\\ =54.94\text{ A}\end{array}$

Conclusion:

Thus the rms current through of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 55 A.

To determine

(c)

The peak value of current in the circuit.

Answer to Problem 81QAP

The peak value of current in the circuit is 77.7 A.

Explanation of Solution

Given:

The rms value of current $i_{\text{rms}}=54.94\text{ A}$

Formula used:

The peak value of current is given by,

  $i_0=\sqrt{2}{i}_{\text{rms}}$

Calculation:

Calculate the value of the peak current by substituting the value of $i_{\text{rms}}$ in the formula.

  $\begin{array}{c}{i}_0=\sqrt{2}{i}_{\text{rms}}\\ =\sqrt{2}\left(54.94\text{ A}\right)\\ =77.7\text{ A}\end{array}$

Conclusion:

Thus the peak value of current in the circuit is 77.7 A.