Question

The voltage across an inductor and the current through the inductor are related by \(v_{\mathrm{L}}=L \Delta i / \Delta t .\) Suppose that $i(t)=I \sin \omega t .\( (a) Write an expression for \)v_{\mathrm{L}}(t) .$ [Hint: Use one of the relationships of Eq. \((20-7) .]\) (b) From your expression for \(v_{\mathrm{L}}(t),\) show that the reactance of the inductor is \(X_{\mathrm{L}}=\omega L_{\mathrm{r}}\) (c) Sketch graphs of \(i(t)\) and \(v_{\mathrm{L}}(t)\) on the same axes. What is the phase difference? Which one leads?

Short answer

Answer: The phase difference between the voltage across an inductor and the current through it is \(\pi/2\) (90 degrees), and the voltage leads the current.

Step-by-step solution

Differentiate the current expression

We have the current expression as $$i(t) = I \sin \omega t$$ Differentiate the given expression to find the change in current with respect to time, $$\frac{di(t)}{dt} = I \cdot \frac{d}{dt}\sin \omega t$$ Using chain rule, $$\frac{di(t)}{dt} = I \cdot \cos \omega t \cdot \frac{d}{dt}(\omega t)$$ $$\frac{di(t)}{dt} = I \omega \cos \omega t$$

Calculate the voltage expression using inductor relation

Using the relation between voltage across an inductor and the current through it, $$v_L(t) = L \frac{\Delta i}{\Delta t}$$ Substitute the change in current with respect to time that we calculated in step 1, $$v_L(t) = L I \omega \cos \omega t$$

Find the reactance of the inductor

Reactance of the inductor is given by, $$X_L = \frac{v_L}{i}$$ Substitute the expressions for \(v_L(t)\) and \(i(t)\), $$X_L = \frac{L I \omega \cos \omega t}{I \sin \omega t}$$ Cancel the current, \(I\), from the equation, $$X_L = \omega L \cdot \frac{\cos \omega t}{\sin \omega t}$$ $$X_L = \omega L_{\mathrm{r}}$$

Sketch graphs for i(t) and v_L(t)

We have the expressions for \(i(t)\) and \(v_L(t)\), $$i(t) = I \sin \omega t$$ $$v_L(t) = L I \omega \cos \omega t$$ Plot the graphs of these expressions on the same set of axes. You will notice that the voltage graph leads the current graph by a phase angle of \(\pi/2\) (90 degrees).

Find the phase difference and which one leads

Compare the forms of the expressions of \(i(t)\) and \(v_{L}(t)\) to note the phase difference, $$i(t) = I \sin \omega t$$ $$v_{L}(t) = L I \omega \cos \omega t$$ The phase difference between the two graphs is \(\pi/2\) (90 degrees), and the voltage \(v_{L}(t)\) leads the current \(i(t)\).