Chapter 29, Problem 27P

(a)

To determine

To Find: The peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity.

Answer to Problem 27P

  $i_{peak}=0.3464\text{A}$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V_1=\left(5\text{ V}\right)\cos \left(\omega t-\alpha \right)$

Potential difference across terminals of one of the other generators, $V_2=\left(5\text{ V}\right)\cos \left(\omega t+\alpha \right)$

Where, $\alpha =\frac{\pi }{6}$

Resistance, $R=25\Omega$

Formula used:

Kirchhoff’s Voltage law:

Sum of the potential in a closed loop circuit is zero.

Calculation:

  

ApplyingKirchhoff’s voltage law:

  $\begin{array}{l}25i=5\cos \left(\omega t-\alpha \right)+5\left(\omega t+\alpha \right)\\ 25i=\left(\cos \left(\omega t-\alpha \right)+\left(\omega t+\alpha \right)\right)\\ 5i=2\cos \omega t\cos \alpha \\ i=\frac{2}{5}\cos \omega t\cos \frac{\pi }{6}\\ i=\frac{\sqrt{3}}{5}\cos \omega t\\ i_{peak}=\frac{\sqrt{3}}{5}\text{A}\\ i_{peak}=0.3464\text{A}\end{array}$

Conclusion:

  $i_{peak}=0.3464\text{A}$ is the peak current in the circuit using Kirchhoff’s loop rule and trigonometric identity

(b)

To determine

To Find: The peak current in the circuit by using a phasor diagram.

Answer to Problem 27P

  $i_{peak}=0.3464\text{A}$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V_1=\left(5\text{ V}\right)\cos \left(\omega t-\alpha \right)$

Potential difference across terminals of one of the other generators, $V_2=\left(5\text{ V}\right)\cos \left(\omega t+\alpha \right)$

Where, $\alpha =\frac{\pi }{6}$

Resistance, $R=25\Omega$

Formula used:

Vector formula

  $\left|\overrightarrow{V}\right|=\sqrt{{\left|{\overrightarrow{V}}_1\right|}^2+{\left|{\overrightarrow{V}}_2\right|}^2+2\left|{\overrightarrow{V}}_1\right|\left|{\overrightarrow{V}}_2\right|\cos 2\alpha }$

Where,

  $\left|\overrightarrow{V}\right|$ is the resultant vector

  ${\overrightarrow{V}}_1$ and ${\overrightarrow{V}}_2$ are the two vectors and $2\alpha$ is the angle between them.

Calculation:

  

  ${\theta }_1$ and ${\theta }_2$ can be given as:

  $\begin{array}{l}{\theta }_1=\omega t-\alpha \\ {\theta }_2=\omega t+\alpha \\ \Rightarrow {\theta }_2-{\theta }_2=\omega t+\alpha -\omega t+\alpha \\ \Rightarrow {\theta }_2-{\theta }_2=2\alpha \end{array}$

Angle between the vectors ${\overrightarrow{V}}_1$ and ${\overrightarrow{V}}_2$ is $2\alpha$

Magnitude of $\overrightarrow{V}$ can be calculated as:

  $\begin{array}{l}\left|\overrightarrow{V}\right|=\sqrt{{\left|{\overrightarrow{V}}_1\right|}^2+{\left|{\overrightarrow{V}}_2\right|}^2+2\left|{\overrightarrow{V}}_1\right|\left|{\overrightarrow{V}}_2\right|\cos 2\alpha }\\ \left|\overrightarrow{V}\right|=\sqrt{5^2+5^2+2\times 5\times 5\cos 2\alpha }\\ \left|\overrightarrow{V}\right|=5\sqrt{2\left(1+\cos 2\alpha \right)}\\ \left|\overrightarrow{V}\right|=5\sqrt{2\times 2{\cos }^2\alpha }\\ \left|\overrightarrow{V}\right|=10\cos \alpha \end{array}$

But,

  $\begin{array}{l}\left|\overrightarrow{V}\right|=I_{peak}R\\ I_{peak}R=10\cos \alpha \\ I_{peak}=\frac{10\cos \alpha }{25}\\ I_{peak}=\frac{2\cos \alpha }{5}\end{array}$

It is given that

  $\alpha =\frac{\pi }{6}$

Therefore,

  $\begin{array}{l}{I}_{peak}=\frac{2}{5}\cos \left(\frac{\pi }{6}\right)\\ I_{peak}=\frac{2}{5}\times \frac{\sqrt{3}}{2}\\ I_{peak}=0.3464\text{A}\end{array}$

Conclusion:

  $i_{peak}=0.3464\text{A}$ is the peak current in the circuit by using a phasor diagram.

(c)

To determine

To Find: The current in the resistor.

Answer to Problem 27P

  $\frac{\sqrt{2}}{50}\left[12\cos \omega t-\sin \omega t\right]$

Explanation of Solution

Given:

Potential difference across terminals of one of the generators, $V_1=\left(5\text{ V}\right)\cos \left(\omega t-\alpha \right)$

Potential difference across terminals of one of the other generators, $V_2=\left(\text{7 V}\right)\cos \left(\omega t+\alpha \right)$

Where, $\alpha =\frac{\pi }{4}$

Resistance, $R=25\Omega$

Formula used:

Kirchhoff’s loop rule

Sum of the potential in the circuit is zero.

Calculation:

ApplyingKirchhoff’s loop rule

  $iR=V_1+V_2$

  $\begin{array}{l}iR=5\cos \left(\omega t-\alpha \right)+7\cos \left(\omega t+\alpha \right)\\ iR=5\cos \omega t\cos \alpha +5\sin \omega t\sin \alpha +7\cos \omega t\cos \alpha -7\sin \omega t\sin \alpha \\ iR=12\cos \omega t\cos \alpha -2\sin \omega t\sin \alpha \end{array}$

For,

  $\begin{array}{l}\alpha =\frac{\pi }{4}\\ \cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\end{array}$

Therefore,

  $\begin{array}{l}i\times 25=\frac{1}{\sqrt{2}}\left[12\cos \omega t-\sin \omega t\right]\\ i=\frac{\sqrt{2}}{50}\left[12\cos \omega t-\sin \omega t\right]\end{array}$

Conclusion:

The current in the resistor is $\frac{\sqrt{2}}{50}\left[12\cos \omega t-\sin \omega t\right]$ .