Chapter 9, Problem 9.1P

To determine

(a)

The value of $V_{ab}(t)$.

Answer to Problem 9.1P

The value of $V_{ab}(t)=-5.32\sin 120\pi t$.

Explanation of Solution

Given Information:

The given circuit is shown below:

$B=0\cdot 2\cos 120\pi tT$

Calculation:

$flux\left(\varphi \right)=\pi {\left(0\cdot 15\right)}^2\beta$

$flux\left(\varphi \right)=1.41\times {10}^{-2}\cos 120\pi t\text{Wb}$

$\begin{array}{c}emf=V_{ba}\left(t\right)=-\frac{d\varphi }{dt}\\ =-\frac{d}{dt}\left(1.41\times {10}^{-2}\cos 120\pi t\text{Wb}\right)\\ =-\left(1.41\times {10}^{-2}\times 120\pi \left(-\sin 120\pi t\right)\right)\\ =5.324\sin 120\pi t\end{array}$

$\begin{array}{c}{V}_{ab}\left(t\right)=-V_{ba}\left(t\right)\\ =-5.32\sin 120\pi t\end{array}$

Conclusion:

The value is $V_{ab}\left(t\right)=-5.32\sin 120\pi t$

To determine

(b)

The value of $I\left(t\right)$.

Answer to Problem 9.1P

The value of $I\left(t\right)$ $=21.3\sin (120\pi t)mA.$

Explanation of Solution

Given Information:

The given circuit is shown below:

$B=0\cdot 2\cos 120\pi tT$

Calculation:

$\begin{array}{c}I\left(t\right)=\frac{V_{ba}\left(t\right)}{R}\\ =\frac{5.32\sin 120\pi t}{250}\\ =21.29\times {10}^{-3}\sin (120\pi t)\text{A}\\ \text{=}21.3\sin (120\pi t)\text{mA}\end{array}$

Conclusion:

The value is, $I\left(t\right)=$

$21.3\sin (120\pi t)\text{mA}$