Chapter 35, Problem 27SP

A coil has an inductance of 0.100 H and a resistance of 12.0 Ω. It is connected to a 110-V, 60.0-Hz line. Determine (a) the reactance of the coil, (b) the impedance of the coil, (c) the current through the coil, (d) the phase angle between current and supply voltage, (e) the power factor of the circuit, and (f) the reading of a wattmeter connected in the circuit.

To determine

The inductive reactance of the coil having inductance $0.100\text{H}$ and a resistance of $12.0\Omega$ when it is connected with $110\text{V, 60Hz}$ line.

Answer to Problem 27SP

Solution:

$37.7\Omega$

Explanation of Solution

Given data:

The value of inductance of the inductor is $0.100\text{ H}$.

The frequency of applied voltage is $60.0\text{ Hz}$.

Formula used:

The expression of inductive reactance of a coil is expressed as,

$X_L=2\pi fL$

Here, $f$ is the frequency of the power supply, and $L$ is the inductance of the inductor.

Explanation:

Recall the expression of inductive reactance.

$X_L=2\pi fL$

Substitute $60\text{ Hz}$ for $f$ and $0.100\text{ H}$ for $L$

$\begin{array}{c}{X}_L=2\pi \left(60\text{ Hz}\right)\left(0.100\text{ H}\right)\\ =37.7\Omega \end{array}$

Conclusion:

The inductive reactanceof the coil is $37.7\Omega$.

To determine

The impedance of the coil having inductance $0.100\text{H}$ and a resistance of $12.0\Omega$ when it is connected with $110\text{V, 60Hz}$ line.

Answer to Problem 27SP

Solution:

$39.6\Omega$

Explanation of Solution

Given data:

The value of inductive reactance of the inductor is $37.7\Omega$.

The resistance of the coil is $12\Omega$.

Formula used:

The impedance of the coil is expressed as,

$Z=\sqrt{R^2+\left(X_L^2\right)}$

Here, $X_L$ is the inductive reactance, and $R$ is the resistance of the coil.

Explanation:

Recall the expression of impedance of the coil,

$Z=\sqrt{\left(X_L^2\right)+R^2}$

Substitute $37.7\Omega$ for $X_L$ and $12\Omega$ for $R$

$\begin{array}{c}Z=\sqrt{{\left(37.7\Omega \right)}^2+{\left(12\Omega \right)}^2}\\ =\sqrt{1421.29+144}\\ \approx 39.6\Omega \end{array}$

Conclusion:

The impedance of the coil is $39.6\Omega$.

To determine

The current through the coil having inductance $0.100\text{H}$ and a resistance of $12.0\Omega$ when it is connected with $110\text{V, 60Hz}$ line.

Answer to Problem 27SP

Solution:

$\text{2}\text{.78}\text{A}$

Explanation of Solution

Given data:

The rms value of voltage is $110\text{ V}$.

The impedance of the coil is $39.6\Omega$.

Formula used:

Write the expression of Ohm’s law for inductor in an ac circuit.

$V_{rms}=I_{rms}Z$

Here, $V_{rms}$ is the rms value of voltage, $I_{rms}$ is the rms value of current, and $Z$ is the impedance of theac circuit.

Explanation:

Recall the expressionof Ohm’s law.

$V_{rms}=I_{rms}Z$

Rearrange for $I_{rms}$.

$I_{rms}=\frac{V_{rms}}{Z}$

Substitute $110\text{ V}$ for $V_{rms}$ and $39.6\Omega$ for $Z$

$\begin{array}{c}{I}_{rms}=\frac{110\text{ V}}{39.6\Omega }\\ =2.78\text{ A}\end{array}$

Conclusion:

The current through the inductor is $\text{2}\text{.78}\text{A}$.

To determine

The phase angle between current and supply voltage when $110\text{V, 60Hz}$ line is connected across the coil having inductance $0.100\text{H}$ and a resistance of $12.0\Omega$.

Answer to Problem 27SP

Solution:

$72.3°$

Explanation of Solution

Given data:

The resistance of the coil is $12\Omega$.

The reactance of the coil is $37.7\Omega$.

Formula used:

The expression for phase angle between voltage and current in RL circuit is expressed as,

$\tan \varphi =\frac{X_L}{R}$

Here, $X_L$ is the inductive reactance, and $R$ is the resistance of the coil.

Explanation:

Recall the expressionfor phase angle between voltage and current.

$\tan \varphi =\frac{X_L}{R}$

Solve for $\varphi$.

$\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)$

Substitute $37.7\Omega$ for $X_L$ and $12\Omega$ for $R$

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{37.7\Omega }{12.0\Omega }\right)\\ =72.3°\end{array}$

Conclusion:

The circuit has inductive in nature; hence the voltage leads by $72.3°$.

To determine

The power factor of the circuithaving the coilof inductance $0.100\text{H}$ and a resistance of $12.0\Omega$ when it is connected with $110\text{V, 60Hz}$ line.

Answer to Problem 27SP

Solution:

$0.303$

Explanation of Solution

Given data:

The phase angle between current and supply voltageis $72.3°$.

Formula used:

The expression of the power factor is expressed as,

$\text{Power factor}=\cos \varphi$

Here, $\varphi$ is the phase angle between current and voltage.

Explanation:

Recall the expression for power factor.

$\text{Power factor}=\cos \varphi$

Substitute $72.3°$ for $\varphi$

$\begin{array}{c}\text{Power factor}=\cos \left(72.3°\right)\\ =0.303\end{array}$

Conclusion:

The power factor of the circuit is $0.303$.

To determine

The reading of wattmeter connected in the circuit across the coil having inductance $0.100\text{H}$ and a resistance of $12.0\Omega$ when it is connected with $110\text{V, 60Hz}$ line.

Answer to Problem 27SP

Solution:

$92.6\text{ W}$

Explanation of Solution

Given data:

The rms value of voltage is $110\text{ V}$.

The rms value of current is $\text{2}\text{.78}\text{A}$.

The power factor of the circuit is $0.303$.

Formula used:

Power consumed in the circuit is calculated as,

$P=VI\cos \varphi$

Here, $V$ is the rms value of voltage, $I$ is the rms value of current, and $\cos \varphi$ is the power factor.

Explanation:

Consider the expression for power consumed in the circuit.

$P=VI\cos \varphi$

Substitute $110\text{ V}$ for $V$, $\text{2}\text{.78}\text{A}$ for $I$, and $0.303$ for $\cos \varphi$

$\begin{array}{c}P=\left(110\text{ V}\right)\left(\text{2}\text{.78}\text{A}\right)\left(0.303\right)\\ =92.6\text{ W}\end{array}$

Conclusion:

The reading of wattmeter connected in the circuit is $92.6\text{ W}$.