Question

a 45 - mh inductor is connected to an ac source of emf with a frequency of 250 hz and a maximum emf of 20 v. if the voltage across the inductor is zero at time t = 0, what is the voltage at time t = 2.0 ms?
-20v
20v
0v
-10v
10v
hint
ensure time is converted appropriately. take the induced emf in the inductor to be ( v_{l}=varepsilon_{m} cos (omega_{d} t-varphi) ) and derive the sign and value of the phase constant ( varphi ) from the fact that the voltage across the inductor is zero at ( t = 0 ).

Explanation:

Step1: Determine the phase constant \(\varphi\)

Given \(V_{L}=\varepsilon_{m}\cos(\omega_{d}t-\varphi)\) and \(V_{L}(0) = 0\).
Substitute \(t = 0\) into the equation: \(0=\varepsilon_{m}\cos(-\varphi)\). Since \(\varepsilon_{m}
eq0\), \(\cos\varphi = 0\). The general solution for \(\cos\varphi=0\) is \(\varphi=\pm\frac{\pi}{2}+ 2n\pi\), \(n\in\mathbb{Z}\). For an inductor - voltage relationship \(v = L\frac{di}{dt}\), and using the standard form of AC voltage \(v = V_{m}\cos(\omega t-\varphi)\), we take \(\varphi=\frac{\pi}{2}\) (by considering the initial - value condition and the nature of inductor voltage - current relationship). So \(V_{L}=\varepsilon_{m}\cos(\omega_{d}t-\frac{\pi}{2})\). Using the trigonometric identity \(\cos(A - B)=\cos A\cos B+\sin A\sin B\), \(V_{L}=\varepsilon_{m}\sin(\omega_{d}t)\) (because \(\cos(\omega_{d}t-\frac{\pi}{2})=\sin(\omega_{d}t)\)).

Step2: Calculate the angular frequency \(\omega_{d}\)

The angular frequency \(\omega_{d}=2\pi f\). Given \(f = 250\space Hz\), so \(\omega_{d}=2\pi\times250=500\pi\space rad/s\).

Step3: Calculate the voltage at \(t = 2.0\space ms=2\times10^{- 3}\space s\)

Substitute \(t = 2\times10^{-3}\space s\), \(\varepsilon_{m}=20\space V\) and \(\omega_{d}=500\pi\space rad/s\) into \(V_{L}=\varepsilon_{m}\sin(\omega_{d}t)\).
\(V_{L}=20\sin(500\pi\times2\times10^{-3})\).
\(500\pi\times2\times10^{-3}=\pi\).
\(\sin(\pi)=0\)

Answer:

\(0V\)